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JEE Mains Maths · Sequences and Series

Telescoping Sums

Sums where each term splits as a difference f(k) − f(k + 1), so almost everything cancels: partial fractions over an AP, quartic denominators, surds, and factorial terms.

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Why this matters

Thirty PYQs. The task is always the same: write the kth term as a difference of two consecutive values of some expression. The skill is seeing which expression. Partial fractions cover half; quartic denominators and surds a third; factorials and functions the rest.

Concept 1 of 3: Partial fractions that cancel in pairs

If each term is f(k)−f(k+1)f(k)-f(k+1), the sum from 1 to nn collapses to f(1)−f(n+1)f(1)-f(n+1). For consecutive AP terms, 1akak+1=1d(1ak−1ak+1)\frac1{a_ka_{k+1}}=\frac1d\left(\frac1{a_k}-\frac1{a_{k+1}}\right). For three factors, 1k(k+1)(k+2)=12[1k(k+1)−1(k+1)(k+2)]\frac1{k(k+1)(k+2)}=\frac12\left[\frac1{k(k+1)}-\frac1{(k+1)(k+2)}\right]: the ½ is the gap between the outer factors.

Definition

  • ∑k=1n[f(k)−f(k+1)]=f(1)−f(n+1)\sum_{k=1}^n[f(k)-f(k+1)]=f(1)-f(n+1).
  • 1akak+1=1d(1ak−1ak+1)\frac1{a_ka_{k+1}}=\frac1d\left(\frac1{a_k}-\frac1{a_{k+1}}\right) for an AP with difference dd.
  • 1k(k+1)(k+2)=12[1k(k+1)−1(k+1)(k+2)]\frac1{k(k+1)(k+2)}=\frac12\left[\frac1{k(k+1)}-\frac1{(k+1)(k+2)}\right].
  • 1k(k+2)=12(1k−1k+2)\frac1{k(k+2)}=\frac12\left(\frac1k-\frac1{k+2}\right): two terms survive at each end.

Telescoping

∑k=1n[f(k)−f(k+1)]=f(1)−f(n+1)\sum_{k=1}^{n}\big[f(k)-f(k+1)\big]=f(1)-f(n+1)

Worked example

Find ∑k=1201k(k+1)\sum_{k=1}^{20}\frac{1}{k(k+1)}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q55Moderate

Example 1 · Sequences and Series · Telescoping Sums

∑n=110(528n(n+1)(n+2))\sum_{n = 1}^{10} \left( \frac{528}{n(n + 1)(n + 2)} \right) is equal to :

A gap of two leaves two terms at each end

In 12(1k−1k+2)\frac12\left(\frac1k-\frac1{k+2}\right) the first two positive terms and the last two negative terms survive, not one of each.
Drill 13 more on partial fractions that cancel in pairs

Concept 2 of 3: Quartic denominators and surds

k4+k2+1=(k2−k+1)(k2+k+1)k^4+k^2+1=(k^2-k+1)(k^2+k+1), and the two factors differ by 2k2k. So kk4+k2+1=12[1k2−k+1−1k2+k+1]\frac{k}{k^4+k^2+1}=\frac12\left[\frac1{k^2-k+1}-\frac1{k^2+k+1}\right], and since k2+k+1k^2+k+1 at kk equals k2−k+1k^2-k+1 at k+1k+1, it telescopes. The same trick handles 4k4+1=(2k2−2k+1)(2k2+2k+1)4k^4+1=(2k^2-2k+1)(2k^2+2k+1). For 1a+b\frac1{\sqrt a+\sqrt b}, rationalise to b−ab−a\frac{\sqrt b-\sqrt a}{b-a}; along an AP every denominator is dd.

Definition

  • k4+k2+1=(k2−k+1)(k2+k+1)k^4+k^2+1=(k^2-k+1)(k^2+k+1).
  • 4k4+1=(2k2−2k+1)(2k2+2k+1)4k^4+1=(2k^2-2k+1)(2k^2+2k+1).
  • 1ak+ak+1=ak+1−akd\frac1{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d} for an AP.

Quartic split

kk4+k2+1=12[1k2−k+1−1k2+k+1]\frac{k}{k^4+k^2+1}=\frac12\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]

Worked example

Find ∑k=15kk4+k2+1\sum_{k=1}^{5}\frac{k}{k^4+k^2+1}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q71Moderate

Example 2 · Sequences and Series · Telescoping Sums

The sum to 10 terms of the series 11+12+14+21+22+24+31+32+34+⋯\frac{1}{1 +1^{2}+1^{4}}+\frac{2}{1 +2^{2}+2^{4}}+\frac{3}{1 +3^{2}+3^{4}}+ \cdots. is

Keep the ½

The two factors differ by 2k2k, not kk, so the split carries a factor 12\frac12. Dropping it doubles the answer, and the doubled value is usually an option.
Drill 10 more on quartic denominators and surds

Concept 3 of 3: Factorials, powers and functions that telescope

k⋅k!=(k+1)!−k!k\cdot k!=(k+1)!-k! and k(k+1)!=1k!−1(k+1)!\frac{k}{(k+1)!}=\frac1{k!}-\frac1{(k+1)!}. More generally r! P(r)r!\,P(r) telescopes when P(r)=(r+1)Q(r+1)−Q(r)P(r)=(r+1)Q(r+1)-Q(r) for a polynomial QQ one degree lower; match coefficients to find QQ. Powers: 1y−1−2y2−1=1y+1\frac1{y-1}-\frac2{y^2-1}=\frac1{y+1} makes 2kx2k+1\frac{2^k}{x^{2^k}+1} telescope. A relation like f(2x)−f(x)=xf(2x)-f(x)=x chains down: f(x)−f(x2n)f(x)-f\left(\frac x{2^n}\right) is a finite GP.

Definition

  • k⋅k!=(k+1)!−k!k\cdot k!=(k+1)!-k!; k(k+1)!=1k!−1(k+1)!\frac k{(k+1)!}=\frac1{k!}-\frac1{(k+1)!}.
  • r! P(r)=(r+1)! Q(r+1)−r! Q(r)r!\,P(r)=(r+1)!\,Q(r+1)-r!\,Q(r) when P(r)=(r+1)Q(r+1)−Q(r)P(r)=(r+1)Q(r+1)-Q(r).
  • 2kx2k+1=2kx2k−1−2k+1x2k+1−1\frac{2^k}{x^{2^k}+1}=\frac{2^k}{x^{2^k}-1}-\frac{2^{k+1}}{x^{2^{k+1}}-1}.

Factorial telescoping

k⋅k!=(k+1)!−k!k\cdot k!=(k+1)!-k!

Worked example

Find ∑k=16k⋅k!\sum_{k=1}^{6}k\cdot k!.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 10 · Q83Moderate

Example 3 · Sequences and Series · Telescoping Sums

If ∑r=110r!(r3+6r2+2r+5)=α(11!)\sum_{r = 1}^{10} r!\left( r^{3}+ 6r^{2}+ 2r + 5 \right)=\alpha(11!), then the value of α\alpha is equal to

The leftover at the lower limit

The sum is Q(n+1)(n+1)!−Q(1)⋅1!Q(n+1)(n+1)!-Q(1)\cdot1!. Q(1)Q(1) is often zero, but not always: work it out before dropping it.
Drill 4 more on factorials, powers and functions that telescope

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