Special sums

A series like 1⋅3+2⋅4+3⋅5+⋯+n(n+2)1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) looks like it needs nn separate additions — until you notice its general term is just a polynomial in nn, and polynomials can be summed term by term using three ready-made formulas. These are the special sums: closed-form expressions for ∑n\sum n, ∑n2\sum n^2 and ∑n3\sum n^3, the totals of the first nn natural numbers, their squares, and their cubes. They sit at the end of the Sequence and Series chapter in ISC Class 11 Algebra, right after arithmetic and geometric progressions, and they are the workhorse tool for summing any series whose nn-th term expands into powers of nn. This article builds the three formulas, proves the trickiest one properly, and drills the split-sum-and-simplify method on a worked example.

The core insight

Any series whose nn-th term is a polynomial in nn can be summed exactly by expanding the term into powers of nn and substituting the three standard formulas ∑n=n(n+1)2\sum n = \frac{n(n+1)}{2}, ∑n2=n(n+1)(2n+1)6\sum n^2 = \frac{n(n+1)(2n+1)}{6} and ∑n3=(n(n+1)2)2\sum n^3 = \left(\frac{n(n+1)}{2}\right)^2.

The core idea

For the first nn natural numbers 1,2,3,…,n1, 2, 3, \ldots, n, three sums recur so often in algebra that they are worth memorising outright. The sum of the numbers themselves:

∑k=1nk=1+2+3+⋯+n=n(n+1)2\sum_{k=1}^{n} k = 1+2+3+\cdots+n = \frac{n(n+1)}{2}

where: nn is the number of terms and the result is the familiar triangular-number formula (Gauss's pairing trick: add the series to itself reversed, giving nn pairs each summing to n+1n+1). The sum of their squares:

∑k=1nk2=12+22+⋯+n2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = 1^2+2^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}

where: the extra factor (2n+1)(2n+1) distinguishes this from the linear sum; it is proved below by telescoping. The sum of their cubes:

∑k=1nk3=13+23+⋯+n3=(n(n+1)2)2=(∑k=1nk)2\sum_{k=1}^{n} k^3 = 1^3+2^3+\cdots+n^3 = \left(\frac{n(n+1)}{2}\right)^2 = \left(\sum_{k=1}^n k\right)^2

where: the right-hand side is just the square of the first formula — a genuinely surprising identity: the sum of the first nn cubes equals the square of the sum of the first nn naturals, not the sum of squares.

Method for a general polynomial series. If the nn-th term TnT_n of a series is a polynomial in nn (say Tn=n2+3nT_n = n^2+3n, or Tn=n(n+1)(n+2)T_n = n(n+1)(n+2) expanded out), proceed in three moves: expand TnT_n into a sum of pure powers of nn; sum term by term using linearity, ∑(aTn+bUn)=a∑Tn+b∑Un\sum(aT_n+bU_n) = a\sum T_n + b\sum U_n, substituting the three formulas above and ∑k=1nc=nc\sum_{k=1}^n c = nc for a constant cc; then factor the result, usually pulling out a common n(n+1)n(n+1) or n6\frac{n}{6}, to present it compactly. The figure below shows why the squares formula carries that awkward-looking (2n+1)(2n+1) factor — it comes from stacking 12,22,…1^2, 2^2, \ldots as nested L-shaped layers, each one wider than the last by an odd amount.

Figure
Nested L shaped layers of unit squares building up the sum of squares one two four nine growing outward from a corner
Sum of squares as nested L shaped layers layer 2: adds 3 squares layer 3: adds 5 squares 1st layer: 1 2nd layer: +3 = 4 3rd layer: +5 = 9 Each new L shaped layer adds 2k−1 unit squares, one more odd number than the last so 1 + 4 + 9 + ... grows as k grows, giving the 2n+1 factor after telescoping running total after n layers equals n(n+1)(2n+1)/6
Derivation AThe sum of squares, derived by telescoping

The goal here is to prove ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} rigorously rather than just quoting it, using nothing beyond expanding a cube and summing a known series — the same telescoping trick that also proves the cube-sum formula, and it settles once and for all why the squares formula carries that extra factor of 2n+12n+1.

Step 1. Start from the algebraic identity (k+1)3−k3=3k2+3k+1(k+1)^3 - k^3 = 3k^2+3k+1, obtained by expanding (k+1)3(k+1)^3 and subtracting k3k^3. This holds for every integer kk, so it holds for k=1,2,…,nk=1,2,\ldots,n in turn.

Step 2. Sum this identity over k=1k=1 to nn. The left side telescopes: writing out the terms, (23−13)+(33−23)+⋯+((n+1)3−n3)(2^3-1^3)+(3^3-2^3)+\cdots+((n+1)^3-n^3), every middle term cancels and only the first and last survive, leaving (n+1)3−13=(n+1)3−1(n+1)^3 - 1^3 = (n+1)^3-1.

Step 3. The right side sums to 3∑k=1nk2+3∑k=1nk+∑k=1n1=3∑k2+3⋅n(n+1)2+n3\sum_{k=1}^n k^2 + 3\sum_{k=1}^n k + \sum_{k=1}^n 1 = 3\sum k^2 + 3\cdot\frac{n(n+1)}{2} + n, using the already-known formula for ∑k\sum k (proved by Gauss pairing above) and ∑k=1n1=n\sum_{k=1}^n 1 = n.

Step 4. Equate the two sides and solve for ∑k2\sum k^2. (n+1)3−1=3∑k2+3n(n+1)2+n(n+1)^3 - 1 = 3\sum k^2 + \frac{3n(n+1)}{2} + n, so 3∑k2=(n+1)3−1−n−3n(n+1)23\sum k^2 = (n+1)^3 - 1 - n - \frac{3n(n+1)}{2}. Expanding (n+1)3=n3+3n2+3n+1(n+1)^3 = n^3+3n^2+3n+1, the right side becomes n3+3n2+2n−3n2+3n2n^3+3n^2+2n - \frac{3n^2+3n}{2}, which combines over a denominator of 22 to 2n3+3n2+n2=n(n+1)(2n+1)2\frac{2n^3+3n^2+n}{2} = \frac{n(n+1)(2n+1)}{2}. So 3∑k2=n(n+1)(2n+1)23\sum k^2 = \frac{n(n+1)(2n+1)}{2}, and dividing by 33 gives ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}, exactly as claimed. Verified: (n+1)3−1(n+1)^3-1 telescopes correctly for n=1,2,3n=1,2,3 (giving 7,26,637,26,63), and 3∑k2+3n(n+1)2+n3\sum k^2 + \frac{3n(n+1)}{2}+n matches each of these when the formula is substituted back in.

Worked example

Find ∑k=1nk(k+2)\displaystyle\sum_{k=1}^{n} k(k+2) in closed form, and evaluate it for n=10n=10.

Step 1. Expand the general term into pure powers of kk: Tk=k(k+2)=k2+2kT_k = k(k+2) = k^2+2k.

Step 2. Split the sum using linearity: ∑k=1nk(k+2)=∑k=1nk2+2∑k=1nk\sum_{k=1}^n k(k+2) = \sum_{k=1}^n k^2 + 2\sum_{k=1}^n k.

Step 3. Substitute the two standard formulas: =n(n+1)(2n+1)6+2⋅n(n+1)2=n(n+1)(2n+1)6+n(n+1)= \frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{6} + n(n+1).

Step 4. Factor out the common n(n+1)n(n+1) and combine over a denominator of 66: =n(n+1)[2n+16+1]=n(n+1)⋅2n+76=n(n+1)(2n+7)6= n(n+1)\left[\frac{2n+1}{6}+1\right] = n(n+1)\cdot\frac{2n+7}{6} = \frac{n(n+1)(2n+7)}{6}.

Step 5. Evaluate at n=10n=10: 10×11×276=29706=495\frac{10\times11\times27}{6} = \frac{2970}{6} = \mathbf{495}. A direct check confirms it: 1(3)+2(4)+⋯+10(12)=3+8+15+24+35+48+63+80+99+120=4951(3)+2(4)+\cdots+10(12) = 3+8+15+24+35+48+63+80+99+120 = 495, matching the formula.

Competitive stretch

For what smallest nn does ∑k=1nk(k+2)\sum_{k=1}^n k(k+2) first exceed 10001000? Using the closed form n(n+1)(2n+7)6\frac{n(n+1)(2n+7)}{6}: at n=12n=12, 12×13×316=48366=806\frac{12\times13\times31}{6} = \frac{4836}{6}=806; at n=13n=13, 13×14×336=60066=1001\frac{13\times14\times33}{6}=\frac{6006}{6}=1001. So the smallest such nn is 13\mathbf{13} — the closed form turns a trial-and-error search into two quick substitutions.

Common traps

  • Applying the formulas to a sum that does not start at k=1k=1. A series starting from k=3k=3, for instance, is (full sum from 11 to nn) minus (sum from 11 to 22), not the raw formula.
  • Forgetting to expand the general term before splitting. Summing k(k+2)k(k+2) "as is" leads nowhere; always expand to k2+2kk^2+2k first, as in Step 1 above.
  • Sign or arithmetic slips while factoring out n(n+1)n(n+1). This is where most errors creep in — always verify the closed form against a small case like n=1n=1 or n=2n=2.
  • Confusing ∑n2\sum n^2 with (∑n)2(\sum n)^2. These differ in general: with n=3n=3, ∑n2=14\sum n^2 = 14 but (∑n)2=36(\sum n)^2 = 36. Only the sum of cubes, not squares, equals the square of the linear sum.

Ask the tutor: Stuck on splitting a series into standard sums, or want your closed-form answer checked against a small case? Ask the AI tutor to work through a special-sums problem with you step by step.

Do I need to memorise the derivation, or just the three formulas?

For exams, the three closed forms — ∑n\sum n, ∑n2\sum n^2, ∑n3\sum n^3 — are what you must recall instantly and apply fast. The telescoping derivation of ∑n2\sum n^2 above is worth understanding once, because it shows the formulas are not arbitrary, and a "prove that" question occasionally does ask for it. But under time pressure in a problem-solving question, quote the formula and move straight to substitution.

Why does the cube sum equal the square of the natural-number sum, but the square sum does not follow the same pattern?

It is a coincidence of the specific exponents, not a general rule connecting sums and their powers. Algebraically, (n(n+1)2)2\left(\frac{n(n+1)}{2}\right)^2 really does simplify to ∑k3\sum k^3 when you multiply it out and compare with the telescoping proof for cubes (built the same way as the squares proof above, using (k+1)4−k4(k+1)^4-k^4). There is no equivalent shortcut linking ∑k2\sum k^2 to (∑k)2(\sum k)^2 — those are simply two different numbers, as the n=3n=3 check in the traps section shows.

How do I handle a series where the general term is a ratio or product of three consecutive terms, like k(k+1)(k+2)?

Expand it fully into powers of kk first — here k(k+1)(k+2)=k3+3k2+2kk(k+1)(k+2) = k^3+3k^2+2k — then split into three separate sums and substitute all three standard formulas (∑k3\sum k^3, ∑k2\sum k^2, ∑k\sum k) at once, exactly as in the worked example but with one extra term. The method never changes, only the amount of algebra; expand, split by linearity, substitute, then factor.

Before this

Where this leads