Summation of Standard Series
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Direct answer
Three power sums anchor nearly every JEE series question: Σk = n(n+1)/2, Σk² = n(n+1)(2n+1)/6, and Σk³ = [n(n+1)/2]² — the last being the square of the first, so the sum of cubes is always a perfect square. Beyond these memorised forms, two techniques finish the rest of the syllabus: telescoping, where a term splits as a difference f(k) − f(k+1) so the middle cancels, and the method of differences for series whose r-th term is itself a polynomial or rational function of r. Arithmetico-geometric progressions (each term = (a + rd)·x^r) yield to the shift-and-subtract manoeuvre, with S∞ = a/(1 − r) + dr/(1 − r)² for |r| < 1.
What you must remember
- Power sums: Σk = n(n+1)/2; Σk² = n(n+1)(2n+1)/6; Σk³ = n²(n+1)²/4; combined forms like Σk(k+1) = n(n+1)(n+2)/3 follow by expansion into power sums.
- Telescoping splits: 1/(n(n+1)) = 1/n − 1/(n+1); more powerfully 1/(n(n+1)(n+2)) = (1/2)[1/(n(n+1)) − 1/((n+1)(n+2))]; Σ 1/(n(n+1)) telescopes to 1 − 1/(n+1).
- Odd and even separations: Σ(2k − 1) = n² (sum of first n odd numbers); Σ of first n even numbers = n(n+1).
- Method of differences: for Σf(k) with f a polynomial of degree d, write the answer as a polynomial of degree d + 1 with unknown coefficients and fit small cases — faster than formal summation by parts.
- AGP finite sum: for Σ(a + (k−1)d)r^(k−1), multiply by r, subtract, solve — the manoeuvre, not the formula, must be automatic.
- AGP infinite sum: a + (a + d)r + (a + 2d)r² + ... = a/(1 − r) + dr/(1 − r)² for |r| < 1; convergence fails otherwise.
- Exponential sums: Σr·x^r and Σr²·x^r for |x| < 1 come from differentiating the geometric sum Σx^r = x/(1 − x) — the calculus route that Advanced prefers.
Telescoping in action
Sum the series 1/(1×2) + 1/(2×3) + ... + 1/(n(n+1)). Each term splits: 1/(k(k+1)) = 1/k − 1/(k+1), so the sum is (1 − 1/2) + (1/2 − 1/3) + ... + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1). Every interior term dies; only the two ends survive. The same machinery handles 1/(4n² − 1): since 1/(4n² − 1) = (1/2)[1/(2n − 1) − 1/(2n + 1)], the partial sum collapses to (1/2)[1 − 1/(2n + 1)], which tends to 1/2.
For a product-of-three rung, Σ 1/(n(n+1)(n+2)) from 1 to n equals (1/4) − 1/(2(n+1)(n+2)) by the two-step split above; the limit 1/4 is a standard numerical-answer answer. The strategic lesson: before touching power sums, always ask whether the general term decomposes. A JEE Main question that looks like heavy algebra is often a two-line telescoping gift, while an honest polynomial series (Σ(3k² + 2k)) is pure power-sum substitution. Diagnosing which world you are in — by staring at the r-th term, not the first three terms — is the entire skill.
Pattern-spotting under time pressure
The classic error is summing by pattern-matching the first few values: the series 1, 3, 6, 10, ... is triangular numbers (n(n+1)/2), but guessing a formula from three terms has no proof and Advanced questions are designed with decoy patterns that break by the fifth term. The second error is forgetting the AGP convergence condition: writing an infinite AGP sum when |r| ≥ 1 hands back a divergent series dressed as a number. A third, quieter trap: Σk³ = (Σk)² invites the false generalisation that Σk³ equals (Σk²)·(Σ1) or similar hybrids — only the square-of-sum identity holds. In numerical-answer format, leave the power-sum expression factorised until the final substitution; expanding n(n+1)(2n+1)/6 with n = 10 as 10·11·21/6 = 385 is both faster and less error-prone than expanding first.
Frequently asked questions
What are the three power sums every JEE aspirant must know?
Σk = n(n+1)/2, Σk² = n(n+1)(2n+1)/6 and Σk³ = [n(n+1)/2]², with the sum of cubes always a perfect square.
How does telescoping evaluate Σ 1/(n(n+1))?
Each term splits as 1/n − 1/(n+1), so all middle terms cancel and the sum from 1 to n is 1 − 1/(n+1) = n/(n+1).
What is the infinite sum of an arithmetico-geometric series?
For |r| < 1, a + (a + d)r + (a + 2d)r² + ... = a/(1 − r) + dr/(1 − r)², obtained by the shift-and-subtract method.
How do you sum Σr·x^r for |x| < 1?
Differentiate the geometric sum Σx^r = x/(1 − x) to get Σr·x^(r−1) = 1/(1 − x)², then multiply by x.
Why must the general term be inspected before choosing a method?
Because the r-th term reveals whether the series telescopes (rational with factorisable denominator), needs power sums (polynomial in r), or is an AGP (linear × exponential) — the first three terms alone can mislead.