Sequences & Series

1.Arithmetic sequences and series

An arithmetic sequence adds the same amount $d$, the common difference, at every step. Its general term is $a_n = a_1 + (n-1)d$. The factor $n-1$, not $n$, is the most common source of off-by-one errors: between the first term and the $n$th term there are exactly $n-1$ gaps, each of size $d$. Equivalently, $a_n$ is a linear function of $n$ with slope $d$, which is why arithmetic sequences are exactly the sequences whose consecutive differences are constant. Two terms $a_m$ and $a_n$ therefore determine the whole sequence, since $a_n - a_m = (n-m)d$.

The sum formula comes from a pairing trick. Write $S = a_1 + a_2 + \cdots + a_n$ and also write the same sum backwards, $S = a_n + a_{n-1} + \cdots + a_1$. Adding the two rows term by term, every column has the same total, because $a_k + a_{n+1-k} = a_1 + a_n$ for each $k$. There are $n$ columns, so $2S = n(a_1 + a_n)$, and therefore $S = \frac{n(a_1+a_n)}{2}$. This is the average of the first and last terms times the number of terms, which is exactly what one expects for evenly spaced numbers. Gauss is said to have found this as a child by pairing $1$ with $100$, $2$ with $99$, and so on; the identity works for any arithmetic sequence, not only consecutive integers.

For a concrete case, consider $2 + 5 + 8 + \cdots + 50$. Here $a_1 = 2$, $d = 3$, and the last term gives $50 = 2 + 3(n-1)$, so $n = 17$. The sum is $\frac{17(2+50)}{2} = 442$. The important point is that the number of terms must be found first, and the solver should be suspicious of any answer that does not come from $n$ explicitly. If the last term is not given, one solves for $n$ from $a_n = a_1 + (n-1)d$ before doing anything else.

Some structural facts recur constantly. Among any odd number of consecutive terms, the middle term equals their average, so the sum of an odd-length block is the middle term times the count. Counting how many terms of an arithmetic sequence lie in an interval is itself a floor problem: the terms $a_1 + kd$ inside an interval correspond to the integers $k$ in an interval of length (end minus start) divided by $d$, so the count is a floor expression. This link to floor functions is developed in the last lesson of this topic.

When a problem describes a list by a rule, first identify $a_1$, $d$, and the last term, then decide which formula is needed. Many problems give two terms and ask for a third or for a sum, so the first skill is to solve the two-equation linear system for $a_1$ and $d$. Remember that $d$ may be negative, and the same formulas work for decreasing sequences. Finally, if a problem sums only part of a sequence, subtract the initial block from the full sum instead of adjusting the formula's index, which is safer under time pressure.

• $a_n = a_1 + (n-1)d$, and $a_n - a_m = (n-m)d$
• $S_n = \frac{n(a_1+a_n)}{2} = \frac{n\,(2a_1+(n-1)d)}{2}$
• $1+2+\cdots+n=\frac{n(n+1)}{2}$, the Gauss pairing identity

• Using $a_1 + nd$ for the $n$th term, which is the $(n+1)$st term.
• Forgetting that the number of terms from $a$ to $b$ with step $d$ is $\frac{b-a}{d}+1$, not $\frac{b-a}{d}$.

2.Geometric sequences and finite sums

A geometric sequence multiplies by a fixed ratio $r$ at each step, so $g_n = a\,r^{\,n-1}$. Arithmetic sequences grow linearly and geometric ones grow or decay exponentially, and confusing the two is one of the most frequent errors on the exam. A quick test: if consecutive differences are constant the sequence is arithmetic, and if consecutive ratios are constant it is geometric. The sequence $3, 6, 12, 24$ has ratio $2$, and a balance growing by $5\%$ a year has ratio $1.05$.

To sum $n$ terms, let $S = a + ar + ar^2 + \cdots + ar^{n-1}$. Multiplying by $r$ shifts every term one place to the right, so $rS = ar + ar^2 + \cdots + ar^{n}$. Subtracting, almost everything cancels: $S - rS = a - ar^{n}$. Hence $S = \frac{a(1-r^n)}{1-r}$ whenever $r \neq 1$. The idea is that a sum and its scaled copy overlap almost entirely, so subtracting them isolates the two ends. This shift-and-subtract method returns in the arithmetico-geometric and recurrence lessons, and it is worth mastering here.

The formula has two useful forms. When $|r|>1$, write it as $S = \frac{a(r^n-1)}{r-1}$ so that all quantities stay positive and sign errors are avoided. When $r=1$ the sum is simply $na$. As an example, the sum of $n$ terms of $2+6+18+\cdots$ is $\frac{2(3^n-1)}{3-1} = 3^n - 1$. So the partial sums of this sequence are one less than the powers of three, a clean identity worth remembering because it solves many counting-by-accumulation problems at once.

Questions of the form 'how many terms until the total exceeds a bound' are solved by evaluating the closed form at small $n$ rather than by logarithms. Since $3^6 - 1 = 728 < 1000 < 2186 = 3^7 - 1$, exactly seven terms are needed to pass $1000$. The exact check is faster and more reliable than an estimate, and it avoids the rounding issues that arise when a logarithm is nearly an integer.

Geometric sums are also the building blocks of many other formulas. The factor $\frac{1}{1-r}$ that appears in the infinite case is already visible in the finite formula, and the next lesson exploits this by letting $n$ grow without bound. Keep in mind that the finite formula needs $r \neq 1$; for $r = 1$ the sum is $na$, and a student who divides by $1-r$ without checking this case will obtain a meaningless expression.

• $S_n = a\,\frac{1-r^n}{1-r}$ for $r \neq 1$, and $S_n = na$ for $r=1$
• $1 + r + r^2 + \cdots + r^{n-1} = \frac{r^n - 1}{r - 1}$
• $2+6+18+\cdots$ with $n$ terms sums to $3^n-1$

• Writing $\frac{a(1-r^n)}{1-r}$ with $n$ terms in one case and $n+1$ in another; the exponent must equal the number of terms.
• Applying the finite formula with $r=1$, where the denominator vanishes; the answer is $na$ instead.

3.Infinite geometric series

An infinite geometric series $a + ar + ar^2 + \cdots$ has a finite sum exactly when its partial sums settle down. From $S_n = a\frac{1-r^n}{1-r}$, the term $r^n$ controls everything. If $|r|<1$ then $r^n \to 0$ and $S_n \to \frac{a}{1-r}$. If $|r|\geq 1$ the powers $r^n$ do not tend to zero, so the partial sums do not converge. For $r=1$ they grow without bound, and for $r=-1$ they oscillate between $a$ and $0$, so no finite sum exists.

Why does $|r|<1$ matter so much? Each term is a fixed fraction of the previous one, so the tail after $n$ terms is itself a geometric series with first term $ar^n$. Its total is $\frac{ar^n}{1-r}$, which is small when $|r|^n$ is small. This gives an error estimate: after $n$ terms one is within $\frac{|a|\,|r|^n}{|1-r|}$ of the total. For $1 + \frac12 + \frac14 + \cdots$ the error after ten terms is $\frac{2}{2^{10}} = \frac{1}{512}$, which is easy to check by hand.

The standard use is converting a repeating decimal into a fraction. Consider $0.\overline{7} = 0.777\ldots = \frac{7}{10} + \frac{7}{100} + \cdots$. The first term is $\frac{7}{10}$ and the ratio is $\frac{1}{10}$, so the sum is $\frac{7/10}{1-1/10} = \frac{7}{9}$. A block $B$ of $p$ repeating digits gives $0.\overline{B} = \frac{B}{10^p-1}$, which is exactly the formula above with ratio $10^{-p}$. This is the fastest route for any repeating decimal, and it is a favorite source of quick points.

The most common error is applying $\frac{a}{1-r}$ without checking the convergence condition. A student who writes $1-2+4-8+\cdots = \frac{1}{1-(-2)} = \frac13$ has used the formula outside its range, because $|r|=2>1$; the series diverges. Always write down $|r|$ before using the formula. A second error concerns the starting index: the formula $\frac{a}{1-r}$ is correct when the first term is $a$, and if the series starts later, say at $k=2$, the first term is still $a$ and the formula still applies, but a solver who starts the sum at a different place must recompute $a$.

The infinite case is the limit of the finite formula and gives a clean principle: a geometric series with ratio $r$ has sum $\frac{a}{1-r}$ exactly when $r$ lies in the open interval $(-1,1)$. Contest problems often hide a convergent geometric series inside a larger expression, such as a sum of $c_k r^k$ where $c_k$ is polynomial in $k$. Those are handled by the arithmetico-geometric method of a later lesson, which reduces the polynomial factor one degree at a time.

• $a + ar + ar^2 + \cdots = \frac{a}{1-r}$ when $|r|<1$
• The tail after $n$ terms equals $\frac{a r^n}{1-r}$
• $0.\overline{B} = \frac{B}{10^p-1}$ for a block $B$ of $p$ repeating digits

• Using $\frac{a}{1-r}$ when $|r| \geq 1$; the series diverges.
• Misreading the first term when the series is written with a sign pattern, such as $1-\frac12+\frac14-\cdots$, whose first term is $1$ and ratio is $-\frac12$.

4.Telescoping sums

A telescoping sum is one whose terms can be written as differences $b_k = f(k) - f(k+1)$ of consecutive values of some function. When the terms are added, every interior value appears once with a plus sign and once with a minus sign and cancels, leaving only the first and last pieces. The partial sum is $\sum_{k=1}^{N} b_k = f(1) - f(N+1)$, and this identity is the whole content of the method. The name comes from the collapse of a telescope.

The main skill is finding $f$. The canonical source is partial fractions. Since $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$, the sum $\sum_{k=1}^{N}\frac{1}{k(k+1)}$ equals $1 - \frac{1}{N+1}$. More generally $\frac{1}{k(k+m)} = \frac{1}{m}\left(\frac{1}{k} - \frac{1}{k+m}\right)$, which is checked by combining the right side over a common denominator. The factor $\frac{1}{m}$ depends on the gap $m$ between the two denominators, and it is a detail that is often lost under time pressure.

To derive the decomposition, write $\frac{1}{k(k+m)} = \frac{A}{k} + \frac{B}{k+m}$. Multiplying by $k(k+m)$ gives $1 = A(k+m) + Bk$. Setting $k=0$ yields $A = \frac{1}{m}$, and setting $k=-m$ yields $B = -\frac{1}{m}$. For a linear factorization two substitutions always suffice, which is why the cover-up method is fast for these problems.

Products telescope as well. In $\prod_{k=2}^{N}\left(1-\frac{1}{k^2}\right)$, each factor equals $\frac{k-1}{k}\cdot\frac{k+1}{k}$. The first factors multiply to $\frac{1}{N}$ and the second to $\frac{N+1}{2}$, so the product is $\frac{N+1}{2N}$. The same idea applies whenever a factor is a ratio $\frac{g(k)}{g(k+1)}$ of consecutive values of some function, since the product then collapses to the first and last values of $g$.

Radicals telescope after rationalizing. Since $(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k}) = 1$, one has $\frac{1}{\sqrt{k}+\sqrt{k+1}} = \sqrt{k+1} - \sqrt{k}$. Summing from $k=1$ to $48$ therefore gives $\sqrt{49} - \sqrt{1} = 6$. The recipe is the same as before: multiply by the conjugate, then look for a function whose values at consecutive integers produce the term.

Two mistakes are common. The first is forgetting the constant $\frac{1}{m}$ when the gap is $m>1$. The second is stopping at a formula in $N$ and forgetting to take the limit or to evaluate at the given $N$. For a convergent telescoping series the sum is $f(1)$ minus the limit of $f$ at infinity, and that limit must be checked, since a limit that does not vanish changes the answer. Also note the sign: the telescope may collapse with the opposite orientation, so compute the first two partial sums to confirm.

• $\sum_{k=1}^{N}\bigl(f(k)-f(k+1)\bigr) = f(1) - f(N+1)$
• $\frac{1}{k(k+m)} = \frac{1}{m}\left(\frac{1}{k} - \frac{1}{k+m}\right)$
• $\frac{1}{\sqrt{k}+\sqrt{k+1}} = \sqrt{k+1} - \sqrt{k}$
• $\sum_{k=1}^{\infty}\frac{1}{k(k+1)} = 1$

• Dropping the factor $\frac{1}{m}$ in $\frac{1}{k(k+m)}$.
• Dropping the boundary term $f(N+1)$ in the partial sum; it vanishes in the limit only when $f(N+1)\to 0$.

5.Arithmetico-geometric series

An arithmetico-geometric series multiplies an arithmetic factor by a geometric one, as in $\sum k\,x^k$ or $\sum (2k+1)\,x^k$. Neither the arithmetic nor the geometric formula applies alone, but the shift-and-subtract idea from the geometric lesson does. The linear coefficient $k$ grows, and the method works by reducing the degree of the polynomial factor by one at each step.

Derive $\sum_{k=1}^{n} k x^k$ carefully. Let $S = x + 2x^2 + \cdots + nx^n$. Then $xS = x^2 + 2x^3 + \cdots + nx^{n+1}$. Subtracting, $(1-x)S = x + x^2 + \cdots + x^n - nx^{n+1}$. The middle part is geometric, so $(1-x)S = \frac{x(1-x^n)}{1-x} - nx^{n+1}$. Solving gives the finite formula, and for $|x|<1$ the term $nx^{n+1}$ tends to zero, so $\sum_{k\geq 1} k x^k = \frac{x}{(1-x)^2}$. Both forms are standard, and the second is the one to memorize.

Check the infinite formula at $x=\frac12$: it gives $\frac{1/2}{(1/2)^2} = 2$, and indeed $\frac12 + \frac24 + \frac38 + \cdots$ sums to $2$. The formula also has a derivative interpretation: $\sum x^k = \frac{x}{1-x}$, and multiplying the derivative by $x$ produces $\sum k x^k$. This explains the squared denominator $(1-x)^{-2}$, though the shift-and-subtract derivation is the one to use under exam conditions.

For a combined example, consider $\sum_{k\geq1}\frac{2k+1}{4^k}$. Split the numerator into $2k$ and $1$. With $x=\frac14$ we have $\sum k x^k = \frac{1/4}{(3/4)^2} = \frac49$ and $\sum x^k = \frac{1/4}{3/4} = \frac13$. Therefore the total is $2\cdot\frac49 + \frac13 = \frac{11}{9}$. Splitting is always legitimate for convergent series, and it is the first move whenever the coefficient is linear in $k$.

The next family is $\sum k^2 x^k = \frac{x(1+x)}{(1-x)^3}$, which follows by applying the same subtraction twice or by differentiating the previous identity. At $x=\frac13$ it gives $\frac{(1/3)(4/3)}{8/27} = \frac32$, and the first few terms $\frac13 + \frac49 + \frac13 + \frac{16}{81} + \cdots$ indeed approach $\frac32$. Expect this family whenever a sum has polynomial terms multiplied by geometric decay.

The common error is a wrong starting index. The formula $\frac{x}{(1-x)^2}$ is for $k$ starting at $1$. If the sum starts at $k=0$, the $k=0$ term of $kx^k$ is $0$, but the plain geometric part gains a constant term $1$, so the total changes. Always rewrite the sum so that the index matches the formula before substituting.

• $\sum_{k\geq 1} k x^k = \frac{x}{(1-x)^2}$ for $|x|<1$
• $\sum_{k\geq 1} k^2 x^k = \frac{x(1+x)}{(1-x)^3}$ for $|x|<1$
• $\sum_{k=1}^{n} k x^k = \frac{x\bigl(1-(n+1)x^n+nx^{n+1}\bigr)}{(1-x)^2}$

• Summing the arithmetic and geometric parts separately with the wrong formulas, for example applying an arithmetic formula to $k x^k$.
• Starting the sum at the wrong index, which changes the constant term in the answer.

6.Linear recurrences and the characteristic equation

A second-order linear recurrence has the form $a_{n+2} = p\,a_{n+1} + q\,a_n$. Such a sequence is determined by $a_1$ and $a_2$, but computing $a_{2025}$ term by term is hopeless, so a closed form is needed. The method is to look for solutions of the form $a_n = \lambda^n$. Substituting gives $\lambda^{n+2} = p\lambda^{n+1} + q\lambda^n$, and dividing by $\lambda^n$ yields the characteristic equation $\lambda^2 = p\lambda + q$.

If the characteristic equation has two distinct roots $\lambda_1 \neq \lambda_2$, then every combination $a_n = A\lambda_1^n + B\lambda_2^n$ satisfies the recurrence. This holds because the recurrence is linear, so any linear combination of solutions is again a solution. The constants $A$ and $B$ are found from the first two terms by solving a $2\times2$ linear system. The logic is the same as for a second-order linear differential equation.

Verify the claim directly. Substituting $A\lambda_1^n + B\lambda_2^n$ into $a_{n+2} - pa_{n+1} - qa_n$ and grouping by $A$ and by $B$ gives $A\lambda_1^n(\lambda_1^2 - p\lambda_1 - q) + B\lambda_2^n(\lambda_2^2 - p\lambda_2 - q)$. Both brackets vanish because each root satisfies the characteristic equation, so the expression is zero for every $n$.

As an example, take $a_{n+2} = 3a_{n+1} - 2a_n$. The characteristic equation is $\lambda^2 - 3\lambda + 2 = (\lambda-1)(\lambda-2) = 0$, so $a_n = A + B\cdot 2^n$. With $a_0 = 0$ and $a_1 = 1$ we get $A + B = 0$ and $A + 2B = 1$, hence $B = 1$, $A = -1$, and $a_n = 2^n - 1$. Check: the recurrence gives $a_2 = 3\cdot 1 - 2\cdot 0 = 3 = 2^2 - 1$. The closed form makes any large index trivial to compute.

A double root $\lambda$ produces a factor of $n$: the general solution is $(A+Bn)\lambda^n$. Complex conjugate roots produce oscillation through cosine and sine, which is recognized by a negative discriminant. On the exam the most common case is distinct real roots, and the main skill is to recognize when the characteristic polynomial factors at once. The Fibonacci recurrence $F_{n+2} = F_{n+1} + F_n$ has roots $\varphi = \frac{1+\sqrt5}{2}$ and $\psi = \frac{1-\sqrt5}{2}$, which gives Binet's formula.

Two warnings. First, put the recurrence into the standard form with the largest index on the left. A recurrence written as $a_n = 3a_{n-1} - 2a_{n-2}$ has the same characteristic equation, but the initial conditions must be matched to the correct indices. Second, a recurrence with a constant term, such as $a_{n+2} = a_{n+1} + a_n + 1$, needs a particular solution added to the homogeneous one, and the constant is found by substituting a constant trial value.

• $a_{n+2} = p\,a_{n+1} + q\,a_n$ has characteristic equation $\lambda^2 = p\lambda + q$
• For distinct roots, $a_n = A\lambda_1^n + B\lambda_2^n$
• For a double root $\lambda$, $a_n = (A + Bn)\lambda^n$
• Binet's formula: $F_n = \frac{\varphi^n - \psi^n}{\sqrt5}$

• Mismatching the initial conditions with the index shift, so the constants $A$ and $B$ are wrong.
• Forgetting the constant particular solution when the recurrence has an additive constant.

7.Periodic sequences and cycles

A sequence is periodic with period $p$ if $a_{n+p} = a_n$ for all $n$. Periodicity is often hidden: a recursive rule does not mention periods, but if the iteration returns to its starting state after $p$ steps the whole sequence repeats from that point on. The practical method is to compute the first several terms, watch for the first repeat of a state, and conclude. For a first-order rule $a_{n+1}=f(a_n)$ the state is one term, and once a value recurs the sequence must cycle from then on.

The argument is short. If $a_k = a_j$ for some $j<k$, then $a_{k+1} = f(a_k) = f(a_j) = a_{j+1}$, and by induction $a_{k+m} = a_{j+m}$ for every $m\geq0$. So the sequence from index $j$ onward has period $k-j$, and any transient before index $j$ is not part of the cycle. The same proof works for a rule that uses the last two terms: the pair $(a_n, a_{n+1})$ is the state, and a repeated pair forces a cycle. The sequence $5, 8, 3, 5, 2, 3, 1, 2, 1, 1, 0, 1, 1, 0, \ldots$ with $a_{n+2}=|a_{n+1}-a_n|$ falls into the block $0,1,1$ only after a transient.

Once the period $p$ is known, reduce indices. Write $N = qp + r$ with $0 \le r < p$. Then $a_N = a_r$ when $r \geq 1$ and $a_N = a_p$ when $r = 0$. For sums, $\sum_{n=1}^{N} a_n = q\sum_{n=1}^{p} a_n + \sum_{n=1}^{r} a_n$. The boundary case $r=0$ means the sum is exactly $q$ full periods, and that is the most common place to make an error.

As an example, let $a_1 = 3$ and $a_{n+1} = 1 - \frac{1}{a_n}$. The terms are $3, \frac23, -\frac12, 3, \ldots$, which is a cycle of length $3$ with sum $3 + \frac23 - \frac12 = \frac{19}{6}$. Since $100 = 3\cdot 33 + 1$, the first $99$ terms are $33$ complete cycles and $a_{100} = a_1 = 3$. The total is $33\cdot\frac{19}{6} + 3 = \frac{215}{2}$. Note that the map $x\mapsto 1-\frac1x$ is a Möbius transformation, and its period is governed by a matrix power, a viewpoint that explains why the period is small here.

Modular cycles are the most common source of periodicity in contests. The powers $2^n \bmod 13$ run through $2,4,8,3,6,12,11,9,5,10,7,1$ and then repeat, so the period is $12$. Any sequence defined by multiplication modulo $m$ is eventually periodic, because there are only finitely many residues. The period is the order of the multiplier, which for $2$ modulo $13$ is $12$ because $2$ is a primitive root modulo $13$. Check the actual cycle rather than guessing the period from the modulus.

Common mistakes include reading the period off the first few terms before the transient has ended, and mixing up whether the first term has index $1$ or $0$, which changes the residue of $N$ by one. Always test your reduction on a small index by listing the terms explicitly, and make sure your formula $a_N = a_r$ agrees with the listed values before you rely on it.

• $a_{n+p} = a_n$ for all $n$ means the sequence has period $p$
• $\sum_{n=1}^{qp+r} a_n = q\sum_{n=1}^{p} a_n + \sum_{n=1}^{r} a_n$ for $0 \le r < p$
• The powers $2^n \bmod 13$ have period $12$

• Taking the period from the first few terms when a transient is still present.
• Confusing the period with the number of residues; the period is the order of the multiplier.

8.Power-sum formulas with derivations

Three power sums come up constantly: $\sum_{k=1}^n k = \frac{n(n+1)}{2}$, $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$, and $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$. These can be memorized, but deriving them is better, because the derivation is a telescoping argument that generalizes to higher powers and to the floor-sum problems of the next lesson.

The first formula follows from the pairing argument of the arithmetic lesson, which gives $2S = n(n+1)$. For squares, start from the telescoping identity $(k+1)^3 - k^3 = 3k^2 + 3k + 1$. Summing over $k = 1, \ldots, n$, the left side telescopes to $(n+1)^3 - 1 = n^3 + 3n^2 + 3n$. The right side is $3\sum k^2 + 3\sum k + n$. Substituting $\sum k = \frac{n(n+1)}{2}$ gives $3\sum k^2 = n^3 + \frac{3n^2 + n}{2} = \frac{n(n+1)(2n+1)}{2}$, so $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$.

The cube formula can be derived from the same telescoping with $(k+1)^4 - k^4$, but a shortcut is more useful on the exam: the cube sum equals the square of the triangular sum, $\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k\right)^2$. A short induction proves it, because if $T_n = \sum_{k\le n} k$ then $(T_n + (n+1))^2 - T_n^2 = 2(n+1)T_n + (n+1)^2 = (n+1)^3$. Check $n=2$: $1^3 + 2^3 = 9 = 3^2$.

The telescoping method extends systematically to any power $\sum k^p$, producing a polynomial of degree $p+1$ whose leading term is $\frac{n^{p+1}}{p+1}$. You can recover that leading term by comparing with the integral $\int_0^n x^p\,dx$, which is a useful sanity check on any answer. Most AMC problems need only the formulas for $p \le 3$, but understanding why they are polynomials of the stated degree prevents mistakes when a problem uses a different power.

In practice, a problem may ask for a combination such as $T - Q$, where $T = \sum k^3$ and $Q = \sum k^2$ over the same range. Compute each piece from its formula rather than adding the terms. Watch for sums that start at a different index: a sum from $k=5$ to $n$ equals the sum from $1$ to $n$ minus the sum from $1$ to $4$. This subtraction is safer than trying to shift the formula's index, and it works equally well for any power.

The most common mistake is confusing the square sum with the cube sum, or using $\sum k^2$ in place of $\sum k$ in the derivation. Test every formula on $n=1$ and $n=2$ before using it. For $n=1$ the three formulas give $1$, $1$, and $1$, and for $n=2$ they give $3$, $5$, and $9$, which catches most errors within a few seconds.

• $\sum_{k=1}^n k = \frac{n(n+1)}{2}$
• $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
• $\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k\right)^2$
• $(k+1)^3 - k^3 = 3k^2 + 3k + 1$

• Confusing $\sum k^2$ with $\left(\sum k\right)^2$, which are different numbers.
• Forgetting that a sum starting at $k=m$ must subtract the first $m-1$ terms.

9.Floor and ceiling sums

The floor $\lfloor x\rfloor$ is the largest integer not exceeding $x$, and the ceiling $\lceil x\rceil$ is the smallest integer not less than $x$. Inside sums, floors turn a continuous quantity into a step function, so the sum becomes a count of lattice points. The basic identity is that $\lfloor x\rfloor = m$ exactly when $m \le x < m+1$. Thus $\lfloor\sqrt{n}\rfloor = k$ exactly when $k^2 \le n < (k+1)^2$, a block of $2k+1$ consecutive integers.

Counting is the cleanest method. For $\sum_{n=1}^{N}\lfloor\sqrt{n}\rfloor$, group the terms by value. The value $k$ appears for exactly $(k+1)^2 - k^2 = 2k+1$ integers $n$, so the sum over a complete range $1 \le n \leq (m+1)^2 - 1$ equals $\sum_{k=1}^{m} k(2k+1)$. Any incomplete final block is handled separately by counting how many integers remain at the last value. This reduces a floor sum to power sums, which is why it follows the power-sum lesson.

For example, to compute $\sum_{n=1}^{100}\lfloor\sqrt{n}\rfloor$, note that $n = 1, \ldots, 99$ form complete blocks for $k=1,\ldots,9$. The contribution is $\sum_{k=1}^{9} k(2k+1) = 2\cdot 285 + 45 = 615$, using $\sum_{k\le 9}k^2 = 285$ and $\sum_{k\le9}k = 45$. The term $n=100$ contributes $10$, giving $625$. The method is reliable because each block is counted exactly once and each count is a simple formula.

Floors of a linear or rational expression are handled by a different idea. The sum $\sum_{n=1}^{N}\lfloor N/n\rfloor$ counts the pairs $(n,j)$ of positive integers with $nj \le N$, since $\lfloor N/n\rfloor$ is the number of $j$ with $nj\le N$. Swapping the roles of $n$ and $j$ gives a symmetry: pairs with $n \le \sqrt{N}$ and pairs with $j \le \sqrt{N}$ overlap in a controlled way. For $N=100$ this yields $2\sum_{n=1}^{10}\lfloor 100/n\rfloor - 100 = 2\cdot 291 - 100 = 482$.

Ceilings relate to floors by $\lceil x\rceil = \lfloor x\rfloor + 1$ for non-integers and $\lceil x\rceil = -\lfloor -x\rfloor$ in general. When a problem mixes them, convert everything to floors first, because counting identities for floors are the standard tools. Be careful with negative arguments: $\lfloor -2.5\rfloor = -3$, not $-2$, and this slip is easy to make when the expression is written with a minus sign inside the floor.

Floor sums are typically evaluated in three steps: identify where the value of the floor changes, count the integers in each constant block with a formula, and add the leftover partial block. Practice with small $N$, where you can list every term, is the quickest route to reliability. Once the method is clear, the same steps handle $\lfloor\sqrt{n}\rfloor$, $\lfloor n/k\rfloor$, and sums of $\lfloor n\alpha\rfloor$ for simple rational $\alpha$.

• $\lfloor x\rfloor = m \iff m \le x < m+1$
• $\#\{n : \lfloor\sqrt{n}\rfloor = k\} = 2k+1$
• $\lceil x\rceil = -\lfloor -x\rfloor$

• Miscounting the length of a block; the number of integers with $\lfloor\sqrt n\rfloor = k$ is $2k+1$, not $2k$.
• Mishandling negative arguments of the floor function.

Example 1: Multiples of seven in a range

Find the sum of all multiples of $7$ from $100$ to $300$, inclusive.

Example 2: Alternating geometric series

Compute $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{3^n}$.

Example 3: Odd-coefficient geometric sum

Compute $\displaystyle\sum_{k=1}^{\infty}\frac{2k-1}{2^k}$.

Example 4: Telescoping over three consecutive integers

Compute $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)}$.