Summation of Series

MathsSequences And SeriesFor JEE aspirants

Summation of series is the technique of finding a closed-form expression for the sum of the first $n$ terms of a given sequence. Standard results include $\sum n = \frac{n ( n + 1 )}{2}$ , $\sum n^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ , and $\sum n^{3} = [\frac{n ( n + 1 )}{2}]^{2}$ . Beyond these, three powerful methods handle almost every JEE problem: expanding the $n$ th term as a polynomial in $n$ , the method of differences when consecutive differences form an AP or GP, and the arithmetico-geometric (AGP) series summation for products of AP and GP terms. Together with telescoping (the $V_{n} - V_{n - 1}$ trick), these cover the full range of series-summation questions in JEE Main and Advanced.

Key Formulas - Quick Reference

$k = 1 \sum n k = \frac{n ( n + 1 )}{2}$
$k = 1 \sum n k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$k = 1 \sum n k^{3} = [\frac{n ( n + 1 )}{2}]^{2}$
$k = 1 \sum n (2 k - 1) = n^{2}$ (first $n$ odd naturals), $k = 1 \sum n 2 k = n (n + 1)$ (first $n$ even naturals)
Infinite geometric: $k = 0 \sum \infty x^{k} = \frac{1}{1 - x}$ , $\sum_{k = 1}^{\infty} k x^{k - 1} = \frac{1}{( 1 - x ) ^{2}}$ for $∣ x ∣ < 1$
AGP sum: $S_{n} = k = 1 \sum n (a + (k - 1) d) b r^{k - 1} = \frac{ab}{1 - r} + \frac{d b r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}} - \frac{( a + ( n - 1 ) d ) b r ^{n}}{1 - r}$
AGP infinite sum ( $∣ r ∣ < 1$ ): $S_{\infty} = \frac{ab}{1 - r} + \frac{d b r}{( 1 - r ) ^{2}}$
Telescoping: $\frac{1}{k ( k + 1 )} = \frac{1}{k} - \frac{1}{k + 1}$ , $\frac{1}{k ( k + 1 ) ( k + 2 )} = \frac{1}{2} [\frac{1}{k ( k + 1 )} - \frac{1}{( k + 1 ) ( k + 2 )}]$

1. Standard Summation Formulas

These closed forms are the foundation for almost every series-summation problem. Every JEE aspirant should have them memorised.

Series	Sum to $n$ terms
$1 + 2 + 3 + \dots + n$	$\frac{n ( n + 1 )}{2}$
$1^{2} + 2^{2} + 3^{2} + \dots + n^{2}$	$\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$1^{3} + 2^{3} + 3^{3} + \dots + n^{3}$	$[\frac{n ( n + 1 )}{2}]^{2}$
$1 + 3 + 5 + \dots + (2 n - 1)$	$n^{2}$
$2 + 4 + 6 + \dots + 2 n$	$n (n + 1)$

Useful identity:

\sum n^{3} = (\sum n)^{2}

. The sum of the first

n

cubes equals the square of the sum of the first

n

naturals. This is a favourite in objective questions.

1.1 Power Series Expansions

For $∣ x ∣ < 1$ , the following infinite series converge to closed forms:

1 + x + x^{2} + x^{3} + \dots = \frac{1}{1 - x}

1 + 2 x + 3 x^{2} + 4 x^{3} + \dots = \frac{1}{( 1 - x ) ^{2}}

These can be derived by termwise differentiation of the geometric series and are essential for infinite AGP and related summations.

2. Summation via the $n$ th Term Expansion

Idea: if the $n$ th term of a series can be written as a polynomial in $n$ (or as a linear combination of $n$ , $n^{2}$ , $n^{3}$ ), then the sum splits by linearity and each piece uses the standard formulas.

Solved Example 1

Find the

n

th term and the sum of

n

terms of the series

1 \cdot 2 \cdot 4 + 2 \cdot 3 \cdot 5 + 3 \cdot 4 \cdot 6 + \dots

Solution:

The $r$ th term is $t_{r} = r (r + 1) (r + 3)$ . Expanding:

t_{r} = r (r + 1) (r + 3) = r^{3} + 4 r^{2} + 3 r

Therefore

S_{n} = r = 1 \sum n t_{r} = \sum r^{3} + 4 \sum r^{2} + 3 \sum r

= [\frac{n ( n + 1 )}{2}]^{2} + 4 \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + 3 \cdot \frac{n ( n + 1 )}{2}

Taking $\frac{n ( n + 1 )}{12}$ common:

S_{n} = \frac{n ( n + 1 )}{12} [3 n (n + 1) + 8 (2 n + 1) + 18] = \frac{n ( n + 1 ) ( 3 n ^{2} + 19 n + 26 )}{12}

3. Method of Differences

Suppose $a_{1}, a_{2}, a_{3}, \dots$ is a sequence such that the successive differences $a_{2} - a_{1}, a_{3} - a_{2}, a_{4} - a_{3}, \dots$ form an AP or a GP. Then the $n$ th term $a_{n}$ can be found by the following trick.

Write the sum $S_{n}$ twice, shifted by one position:

S = a_{1} + a_{2} + a_{3} + \dots + a_{n - 1} + a_{n}

S = a_{1} + a_{1} + a_{2} + \dots + a_{n - 2} + a_{n - 1} + a_{n}

Subtracting (aligning $a_{n}$ with $a_{n}$ in the second line and shifting):

a_{n} = a_{1} + [(a_{2} - a_{1}) + (a_{3} - a_{2}) + \dots + (a_{n} - a_{n - 1})]

The bracketed differences form an AP or GP (by assumption), so their sum is known. Once $a_{n}$ is expressed as a polynomial or geometric expression in $n$ , the full sum $S_{n} = \sum a_{r}$ follows from the standard formulas.

Solved Example 2

Find the sum of

n

terms of the series

1 + 5 + 11 + 19 + 29 + \dots

Solution:

Successive differences: $4, 6, 8, 10, \dots$ These are in AP (first term $4$ , common difference $2$ ).

By the method of differences:

t_{n} = 1 + [4 + 6 + 8 + \dots up to (n - 1) terms]

= 1 + \frac{n - 1}{2} [2 (4) + (n - 2) (2)] = 1 + (n - 1) (n + 2) = n^{2} + n - 1

Now sum:

S_{n} = r = 1 \sum n (r^{2} + r - 1) = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + \frac{n ( n + 1 )}{2} - n

= \frac{n ( n + 1 )}{2} [\frac{2 n + 1}{3} + 1] - n = \frac{n ( n + 1 ) ( 2 n + 4 )}{6} - n = \frac{n ( n + 1 ) ( n + 2 )}{3} - n

= \frac{n ( n ^{2} + 3 n - 1 )}{3}

Solved Example 3

Find the sum of the first

n

terms of the series

5 + 7 + 11 + 17 + 25 + \dots

Solution:

Successive differences: $2, 4, 6, 8, \dots$ These are in AP.

By the method of differences, the $r$ th term is

t_{r} = 5 + [2 + 4 + 6 + \dots up to (r - 1) terms] = 5 + 2 \cdot \frac{( r - 1 ) r}{2} = r^{2} - r + 5

Summing:

S_{n} = r = 1 \sum n r^{2} - r = 1 \sum n r + 5 r = 1 \sum n 1 = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{n ( n + 1 )}{2} + 5 n

= \frac{n}{6} [(n + 1) (2 n + 1) - 3 (n + 1) + 30] = \frac{n}{3} (n^{2} + 14)

4. Arithmetico-Geometric Progression (AGP)

An arithmetico-geometric progression is a sequence formed by multiplying corresponding terms of an AP and a GP. If

a_{1}, a_{2}, a_{3}, \dots

is an AP and

b_{1}, b_{2}, b_{3}, \dots

is a GP, then

a_{1} b_{1}, a_{2} b_{2}, a_{3} b_{3}, \dots

is an AGP. In standard form, the

k

th term is

(a + (k - 1) d) b r^{k - 1}

4.1 Sum of $n$ Terms of an AGP

Consider

S_{n} = ab + (a + d) b r + (a + 2 d) b r^{2} + \dots + (a + (n - 1) d) b r^{n - 1}

Multiply both sides by $r$ :

r S_{n} = ab r + (a + d) b r^{2} + \dots + (a + (n - 2) d) b r^{n - 1} + (a + (n - 1) d) b r^{n}

Subtracting the second from the first, the "shift" causes all middle terms to line up in a GP:

(1 - r) S_{n} = ab + d b r + d b r^{2} + \dots + d b r^{n - 1} - (a + (n - 1) d) b r^{n}

The middle part is a finite GP of $n - 1$ terms with first term $d b r$ and common ratio $r$ :

(1 - r) S_{n} = ab + \frac{d b r ( 1 - r ^{n - 1} )}{1 - r} - (a + (n - 1) d) b r^{n}

S_{n} = \frac{ab}{1 - r} + \frac{d b r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}} - \frac{( a + ( n - 1 ) d ) b r ^{n}}{1 - r}

4.2 Infinite AGP Sum

If $∣ r ∣ < 1$ , then $r^{n} \to 0$ as $n \to \infty$ , and

S_{\infty} = \frac{ab}{1 - r} + \frac{d b r}{( 1 - r ) ^{2}}

Practical tip: memorising the boxed formula is optional. What is essential is the multiply by $r$ and subtract technique, because you will reuse it on any series that mixes AP-linear and GP-exponential factors, even variants the boxed formula does not directly cover.

Solved Example 4

Find the sum of the series

1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \dots + 100 \cdot 2^{100}

Solution:

Let $S = 1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \dots + 100 \cdot 2^{100}$ .

Multiply by $2$ :

2 S = 1 \cdot 2^{2} + 2 \cdot 2^{3} + \dots + 99 \cdot 2^{100} + 100 \cdot 2^{101}

Subtract $2 S$ from $S$ :

- S = 1 \cdot 2 + 1 \cdot 2^{2} + 1 \cdot 2^{3} + \dots + 1 \cdot 2^{100} - 100 \cdot 2^{101}

The first part is a GP with $a = 2, r = 2$ , 100 terms:

- S = 2 \cdot \frac{2 ^{100} - 1}{2 - 1} - 100 \cdot 2^{101} = 2^{101} - 2 - 100 \cdot 2^{101}

\Rightarrow S = 100 \cdot 2^{101} - 2^{101} + 2 = 99 \cdot 2^{101} + 2

Solved Example 5

Find the sum

1 + 2 \cdot 2 + 3 \cdot 2^{2} + 4 \cdot 2^{3} + \dots + 100 \cdot 2^{99}

Solution:

Let $S_{n} = 1 + 2 \cdot 2 + 3 \cdot 2^{2} + \dots + t_{n}$ , where $t_{n} = n \cdot 2^{n - 1}$ and $n = 100$ here.

Multiply by $2$ :

2 S_{n} = 1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \dots + (n - 1) \cdot 2^{n - 1} + n \cdot 2^{n}

Subtract:

S_{n} - 2 S_{n} = 1 + [(2 - 1) \cdot 2 + (3 - 2) \cdot 2^{2} + \dots + (n - (n - 1)) \cdot 2^{n - 1}] - n \cdot 2^{n}

- S_{n} = 1 + [2 + 2^{2} + \dots + 2^{n - 1}] - n \cdot 2^{n}

The bracketed GP: $2 + 2^{2} + \dots + 2^{n - 1} = 2 \cdot \frac{2 ^{n - 1} - 1}{2 - 1} = 2^{n} - 2$ . So

- S_{n} = 1 + 2^{n} - 2 - n \cdot 2^{n} = - 1 + 2^{n} (1 - n)

\Rightarrow S_{n} = 1 - (1 - n) \cdot 2^{n} = 1 + (n - 1) \cdot 2^{n}

For $n = 100$ : $S = 1 + 99 \cdot 2^{100}$ .

5. Telescoping and the $V_{n} - V_{n - 1}$ Method

A telescoping series is one where each term can be expressed as a difference $V_{k} - V_{k - 1}$ (or $V_{k} - V_{k + 1}$ ) for some auxiliary sequence $V_{k}$ . Adding the series then collapses everything except boundary terms.

k = 1 \sum n (V_{k} - V_{k - 1}) = V_{n} - V_{0}

5.1 Standard Telescoping Identities

These identities appear repeatedly in JEE problems. Memorise them.

Identity	Sum to $n$ terms
$\frac{1}{k ( k + 1 )} = \frac{1}{k} - \frac{1}{k + 1}$	$k = 1 \sum n \frac{1}{k ( k + 1 )} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$
$\frac{1}{k ( k + 1 ) ( k + 2 )} = \frac{1}{2} [\frac{1}{k ( k + 1 )} - \frac{1}{( k + 1 ) ( k + 2 )}]$	$k = 1 \sum n \frac{1}{k ( k + 1 ) ( k + 2 )} = \frac{1}{2} [\frac{1}{2} - \frac{1}{( n + 1 ) ( n + 2 )}]$
$k \cdot k! = (k + 1)! - k!$	$k = 1 \sum n k \cdot k! = (n + 1)! - 1$
$\frac{1}{k ^{2} - 1} = \frac{1}{2} [\frac{1}{k - 1} - \frac{1}{k + 1}]$	$k = 2 \sum n \frac{1}{k ^{2} - 1} = \frac{1}{2} [\frac{3}{2} - \frac{1}{n} - \frac{1}{n + 1}]$

5.2 The $V_{n}$ Method for Product-Type Series

For series whose $n$ th term is a product of consecutive integers (like $t_{r} = r (r + 1) (r + 2)$ or $t_{r} = 1/ [r (r + 1) (r + 2)]$ ), the $V_{n}$ method extends the pattern:

If $t_{r} = f (r) \cdot g (r) \cdot h (r)$ or its reciprocal, define $V_{r}$ so that $t_{r} = V_{r} - V_{r - 1}$ . For $t_{r} = r (r + 1) (r + 2)$ , take $V_{r} = \frac{r ( r + 1 ) ( r + 2 ) ( r + 3 )}{4}$ . Verify:

V_{r} - V_{r - 1} = \frac{r ( r + 1 ) ( r + 2 ) ( r + 3 ) - ( r - 1 ) r ( r + 1 ) ( r + 2 )}{4} = \frac{r ( r + 1 ) ( r + 2 ) [( r + 3 ) - ( r - 1 )]}{4} = r (r + 1) (r + 2)

So $r = 1 \sum n r (r + 1) (r + 2) = V_{n} - V_{0} = \frac{n ( n + 1 ) ( n + 2 ) ( n + 3 )}{4}$ .

The pattern generalises: for a product of $m$ consecutive integers starting at $r$ , sum $= \frac{r ( r + 1 ) \dots ( r + m )}{m + 1}$ evaluated at the boundaries.

Common Mistakes to Avoid

Watch out

Applying $\sum n^{3} = (\sum n)^{2}$ blindly. This is only true for the sum of cubes of the first $n$ naturals, not for any cubed sequence.
Forgetting the $∣ r ∣ < 1$ condition for infinite GP or infinite AGP sums. Both formulas diverge otherwise. A common trap in objective questions.
Multiplying by $r$ wrong in AGP. When you write $r S_{n}$ , remember that it has one fewer term aligned with $S_{n}$ ; the last term of $r S_{n}$ hangs off the right, and the first term of $S_{n}$ hangs off the left. Subtract carefully.
Assuming successive differences always form an AP. Sometimes they form a GP (rate of growth is exponential). Check the second differences before applying the AP method of differences.
Missing partial fraction constants. When writing $\frac{1}{k ( k + 1 ) ( k + 2 )}$ as a telescoping difference, the factor of $1/2$ is essential. Always verify by expanding one term.
Off-by-one errors in the $V_{n}$ method. Always compute $V_{0}$ (or the appropriate boundary) explicitly rather than assuming it is zero. For $V_{r} = r (r + 1) (r + 2) (r + 3) /4$ , $V_{0} = 0$ , but this is not universal.

Frequently Asked Questions

Q1. What is an arithmetico-geometric progression?

An AGP is a sequence whose $k$ th term is the product of the $k$ th term of an AP and the $k$ th term of a GP. In standard form, term $k$ is $(a + (k - 1) d) \cdot b r^{k - 1}$ . Its sum is found by multiplying $S$ by $r$ and subtracting, which converts the middle into a GP that can be summed in closed form.

Q2. When should I use the method of differences?

Use the method of differences when the terms of a series do not fit a standard progression, but the successive differences $a_{2} - a_{1}, a_{3} - a_{2}, \dots$ form an AP or a GP. Recognise this by inspecting the first two or three differences of the given series before trying anything else.

Q3. What is the shortcut for $1 + 2 + 3 + \dots + n$ and its higher-power versions?

The formulas are $\sum n = \frac{n ( n + 1 )}{2}$ , $\sum n^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ , and $\sum n^{3} = [\frac{n ( n + 1 )}{2}]^{2}$ . These three are the workhorses of series summation and should be memorised. Higher powers ( $\sum n^{4}, \sum n^{5}$ ) are derived on-the-fly by the same techniques when needed.

Q4. What is telescoping and how do I spot it?

Telescoping is when each term of a series can be written as a difference $V_{k} - V_{k - 1}$ , so the sum collapses to the boundary values. Spot it when the general term has the form $\frac{1}{k ( k + 1 )}$ , $\frac{1}{k ( k + 1 ) ( k + 2 )}$ , $k \cdot k!$ , or any product of consecutive integers or their reciprocals. Partial fractions almost always reveal a telescoping structure.

Q5. How is the AGP formula derived?

Write the sum $S_{n}$ and multiply by the GP ratio $r$ to get $r S_{n}$ . Shift $r S_{n}$ by one position and subtract from $S_{n}$ . The AP coefficient parts telescope: instead of $a, a + d, a + 2 d, \dots$ multiplying each term, only constant differences $d$ remain in the middle. The result is one initial term, a finite GP in the middle, and one final term. Solve for $S_{n}$ from $(1 - r) S_{n} = \dots$

Q6. How do I sum series like $\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots$ ?

Use the telescoping identity $\frac{1}{k ( k + 1 ) ( k + 2 )} = \frac{1}{2} [\frac{1}{k ( k + 1 )} - \frac{1}{( k + 1 ) ( k + 2 )}]$ . Summing from $k = 1$ to $n$ makes middle terms cancel, leaving only the first and last terms: $\frac{1}{2} [\frac{1}{2} - \frac{1}{( n + 1 ) ( n + 2 )}]$ , which approaches $\frac{1}{4}$ as $n \to \infty$ .

Q7. Is there a formula for $\sum n \cdot 2^{n}$ or similar mixed series?

Yes. Any series of the form $\sum (polynomial in n) \cdot r^{n}$ is an AGP variant. Apply the multiply-by- $r$ and subtract method. For $\sum_{k = 1}^{n} k \cdot 2^{k}$ , this gives $S_{n} = (n - 1) \cdot 2^{n + 1} + 2$ . General polynomial-times-geometric series follow the same procedure with more algebra.

Q8. Which summation techniques are most tested in JEE Advanced?

JEE Advanced favours (i) telescoping with partial fractions, (ii) AGP with unusual polynomial factors, and (iii) the $V_{n}$ method for products of consecutive integers. Straight applications of $\sum n, \sum n^{2}, \sum n^{3}$ are more common in JEE Main. Advanced usually combines two techniques in one problem, so master each in isolation first.

Previous year questions on Summation of Series

32 questions from past papers, each with a step-by-step solution.

Show all 32 questions

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1. Standard Summation Formulas

1.1 Power Series Expansions

2. Summation via the nth Term Expansion

3. Method of Differences

4. Arithmetico-Geometric Progression (AGP)

4.1 Sum of n Terms of an AGP

4.2 Infinite AGP Sum

5. Telescoping and the Vn​−Vn−1​ Method

5.1 Standard Telescoping Identities

5.2 The Vn​ Method for Product-Type Series

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is an arithmetico-geometric progression?

Q2. When should I use the method of differences?

Q3. What is the shortcut for 1+2+3+⋯+n and its higher-power versions?

Q4. What is telescoping and how do I spot it?

Q5. How is the AGP formula derived?

Q6. How do I sum series like 1⋅2⋅31​+2⋅3⋅41​+⋯?

Q7. Is there a formula for ∑n⋅2n or similar mixed series?

Q8. Which summation techniques are most tested in JEE Advanced?

Previous year questions on Summation of Series

Ready to master Sequences And Series?

2. Summation via the $n$ th Term Expansion

4.1 Sum of $n$ Terms of an AGP

5. Telescoping and the $V_{n} - V_{n - 1}$ Method

5.2 The $V_{n}$ Method for Product-Type Series

Q3. What is the shortcut for $1 + 2 + 3 + \dots + n$ and its higher-power versions?

Q6. How do I sum series like $\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots$ ?

Q7. Is there a formula for $\sum n \cdot 2^{n}$ or similar mixed series?