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JEE Mains Maths · Sequences and Series

Arithmetico-Geometric and Exponential Series

Series whose terms are a polynomial times a power, summed by multiplying by the ratio and subtracting, and series with factorials or k in the denominator, summed through e and the logarithm series.

All Sequences and Series notesJEE Mains Maths strategy

Why this matters

Twenty-three PYQs. Two thirds are arithmetico-geometric: numerators in AP, or with differences in AP, over powers of one number. The rest divide by n! or by n and are summed through e or the logarithm series. Three ideas cover the page.

Concept 1 of 3: Multiply by the ratio and subtract

In ∑(a+(k−1)d)rk−1\sum\big(a+(k-1)d\big)r^{k-1}, multiply the whole series by rr and subtract it from the original. Matching powers line up, the numerators differ by dd every time, and a plain GP is left. For ∣r∣<1|r|<1 the infinite sum is a1−r+dr(1−r)2\frac a{1-r}+\frac{dr}{(1-r)^2}.

Definition

  • S−rS=a+d(r+r2+… )S-rS=a+d(r+r^2+\dots) minus the last term (finite case).
  • Infinite, ∣r∣<1|r|<1: S=a1−r+dr(1−r)2S=\frac a{1-r}+\frac{dr}{(1-r)^2}.
  • ∑k≥1kxk−1=1(1−x)2\sum_{k\ge1}kx^{k-1}=\frac1{(1-x)^2}; ∑k≥1kxk=x(1−x)2\sum_{k\ge1}kx^k=\frac x{(1-x)^2}.
  • Finite: ∑k=1nk 2k−1=(n−1)2n+1\sum_{k=1}^nk\,2^{k-1}=(n-1)2^n+1.

Infinite AGP

S∞=a1−r+dr(1−r)2S_\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}

Worked example

Find 1+32+54+78+…1+\frac32+\frac54+\frac78+\dots.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 June 2022 · Q158Moderate

Example 1 · Sequences and Series · Arithmetico-Geometric and Exponential Series

The sum 1+2⋅3+3⋅32+…..+10⋅391 + 2 \cdot 3 + 3 \cdot3^{2}+ \ldots.. + 10 \cdot3^{9} is equal to

The finite case has a last term

For a finite AGP, the subtraction leaves −(a+(n−1)d)rn-\big(a+(n-1)d\big)r^n at the end. Dropping it gives the infinite formula, which is wrong here.
Drill 8 more on multiply by the ratio and subtract

Concept 2 of 3: Numerators whose differences are in AP

When the numerators' differences are themselves in AP, as in 1,4,8,13,191,4,8,13,19 (differences 3,4,5,63,4,5,6), one subtraction gives an arithmetico-geometric series, and a second gives a GP. Numerators k(k+1)k(k+1) or k(k+1)2\frac{k(k+1)}2 have closed forms from differentiating the geometric series twice.

Definition

  • Subtract twice for a quadratic numerator: S(1−r)2S(1-r)^2 leaves a GP.
  • ∑k≥1k(k+1)2xk−1=1(1−x)3\sum_{k\ge1}\frac{k(k+1)}2x^{k-1}=\frac1{(1-x)^3}.
  • ∑k≥1k(k+1)xk−1=2(1−x)3\sum_{k\ge1}k(k+1)x^{k-1}=\frac2{(1-x)^3}.
  • ∑k≥1k2xk−1=1+x(1−x)3\sum_{k\ge1}k^2x^{k-1}=\frac{1+x}{(1-x)^3}.

Quadratic numerators

∑k≥1k(k+1) xk−1=2(1−x)3\sum_{k\ge1}k(k+1)\,x^{k-1}=\frac{2}{(1-x)^3}

Worked example

Find 1+3x+6x2+10x3+…1+3x+6x^2+10x^3+\dots at x=12x=\frac12.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 June 2022 · Q155Moderate

Example 2 · Sequences and Series · Arithmetico-Geometric and Exponential Series

Let S=2+67+1272+2073+3074+…S = 2 +\frac{6}{7}+\frac{12}{7^{2}}+\frac{20}{7^{3}}+\frac{30}{7^{4}}+ \ldots. then 4S4S is equal to

One subtraction is not enough

After the first subtraction the numerators are in AP, so it is still an arithmetico-geometric series. Subtract again before using the GP formula.
Drill 5 more on numerators whose differences are in ap

Concept 3 of 3: Series summed through e and the logarithm series

e=∑n≥01n!e=\sum_{n\ge0}\frac1{n!} and 1e=∑n≥0(−1)nn!\frac1e=\sum_{n\ge0}\frac{(-1)^n}{n!}. For ∑P(n)n!\sum\frac{P(n)}{n!}, rewrite P(n)P(n) in the pieces n(n−1)n(n-1), nn, 11: each piece cancels against the factorial and leaves another copy of ee. Only even or only odd factorials give 12(e±1e)\frac12\left(e\pm\frac1e\right). For nn in the denominator use −ln⁡(1−x)=x+x22+x33+…-\ln(1-x)=x+\frac{x^2}2+\frac{x^3}3+\dots; a term like k+1kxk\frac{k+1}{k}x^k splits into a GP plus this series.

Definition

  • ∑n≥0n(n−1)n!=∑n≥0nn!=∑n≥01n!=e\sum_{n\ge0}\frac{n(n-1)}{n!}=\sum_{n\ge0}\frac{n}{n!}=\sum_{n\ge0}\frac1{n!}=e.
  • ∑n≥01(2n)!=12(e+1e)\sum_{n\ge0}\frac1{(2n)!}=\frac12\left(e+\frac1e\right); ∑n≥01(2n+1)!=12(e−1e)\sum_{n\ge0}\frac1{(2n+1)!}=\frac12\left(e-\frac1e\right).
  • ∑n≥1xnn=−ln⁡(1−x)\sum_{n\ge1}\frac{x^n}n=-\ln(1-x), ∣x∣<1|x|<1.

The exponential pieces

∑n≥0n(n−1)n!=∑n≥0nn!=e\sum_{n\ge0}\frac{n(n-1)}{n!}=\sum_{n\ge0}\frac{n}{n!}=e

Worked example

Find ∑n=0∞n2+1n!\sum_{n=0}^{\infty}\frac{n^2+1}{n!}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q152Moderate

Example 3 · Sequences and Series · Arithmetico-Geometric and Exponential Series

The sum ∑n=1∞2n2+3n+4(2n)!\sum_{n = 1}^{\infty} \frac{2n^{2}+ 3n + 4}{(2n)!} is equal to:

Watch where the sum starts

A sum from n=1n=1 leaves out the n=0n=0 term of the ee series. Subtract that term, or the answer is off by a constant.
Drill 7 more on series summed through e and the logarithm series

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Watch out for (3)

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