JEE Mains Maths · Sequences and Series

Sums by Standard Formulas

Adding series whose kth term is a polynomial in k: the formulas for Σk, Σk² and Σk³, finding the kth term from differences, and alternating, grouped and greatest-integer sums.

All Sequences and Series notes JEE Mains Maths strategy

Why this matters

Thirty-four PYQs. Each one adds a series that is not a progression. The work is to find the kth term, from the pattern, from the differences, or from a given sum, and then add it with the three standard formulas. Three ideas cover the page.

Concept 1 of 3: Σk, Σk² and Σk³ applied to a polynomial kth term

Write the

k

th term as a polynomial in

k

, expand, and add each power with its formula. When each term is itself a sum, such as

1^2+2^2+\dots+k^2

, put in its formula first; it often cancels against the rest of the term.

Definition

$\sum_{k=1}^n k=\frac{n(n+1)}2$ .
$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6$ .
$\sum_{k=1}^n k^3=\left[\frac{n(n+1)}2\right]^2$ .
$\sum_{k=1}^n 1=n$ .

Sum of cubes

\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2

Worked example

Find

\sum_{k=1}^{10}k(k+1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q56Moderate

Example 1 · Sequences and Series · Sums by Standard Formulas

The sum

1 +\frac{1}{2}\left( 1^{2}+2^{2} \right)+\frac{1}{3}\left( 1^{2}+2^{2}+3^{2} \right)+ ...

up to 10 terms is equal to :

Sum first, then put in the limit

Expand the term in

k

, add over

k

, and only then put in the upper limit. Putting

n

inside the term too early mixes the running index with the limit.

Drill 13 more on σk, σk² and σk³ applied to a polynomial kth term

Concept 2 of 3: Finding the kth term from differences or from the sum

When the gaps between terms are themselves in AP (the second differences are constant), the

k

th term is a quadratic

ak^2+bk+c

; fit it from the first three terms. When the sum

S_n

is given, the terms are

a_n=S_n-S_{n-1}

. A recurrence such as

a_{n+2}-2a_{n+1}+a_n=1

says the second differences are constant, so it too gives a quadratic.

Definition

Constant second differences: $T_k=ak^2+bk+c$ , with $2a$ the second difference.
Sum given: $a_n=S_n-S_{n-1}$ .
Fit from three terms, then check against the fourth.

Quadratic kth term

T_k=ak^2+bk+c,\qquad 2a=\text{second difference}

Worked example

Add the first 10 terms of

3+7+13+21+\dots

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q68Moderate

Example 2 · Sequences and Series · Sums by Standard Formulas

The sum of the first 20 terms of the series

5 + 11 +

19 + 29 + 41 + \ldots

Check the fourth term

Three terms always fit some quadratic. Check the fitted

T_k

against a fourth term before adding; if it fails, the differences are not in AP.

Drill 8 more on finding the kth term from differences or from the sum

Concept 3 of 3: Alternating, grouped and greatest-integer sums

An alternating sum is (all terms) minus twice (the negative ones), or a sum of consecutive pairs. When the

k

th group holds

k

numbers, group

k

ends at

1+2+\dots+k

; when it holds

2k-1

, it ends at

k^2

. A greatest-integer sum counts how often each value occurs:

[\sqrt k]=m

for the

2m+1

values

m^2\le k<(m+1)^2

. A double sum of

\min\{i,j\}

counts how many cells hold each value.

Definition

Alternating: $\sum(-1)^{k+1}t_k=\sum t_k-2\sum_{k\text{ even}}t_k$ .
$1^2-2^2+3^2-\dots-(2n)^2=-n(2n+1)$ .
Groups of size $k$ : group $k$ ends at $\frac{k(k+1)}2$ . Size $2k-1$ : ends at $k^2$ .
$[\sqrt k]=m$ for $2m+1$ values of $k$ .

Alternating squares

1^2-2^2+3^2-\dots-(2n)^2=-n(2n+1)

Worked example

Find

1^2-2^2+3^2-4^2+\dots+19^2-20^2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q55Moderate

Example 3 · Sequences and Series · Sums by Standard Formulas

The value of

1^{3}-2^{3}+3^{3}- \ldots. +15^{3}

is:

An odd count leaves one term unpaired

Pairing works cleanly only for an even number of terms. With an odd count, pair the rest and add the last term separately.

Drill 10 more on alternating, grouped and greatest-integer sums

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)

Σk, Σk² and Σk³ applied to a polynomial kth term
Sum of cubes
$\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$
Finding the kth term from differences or from the sum
Quadratic kth term
$T_k=ak^2+bk+c,\qquad 2a=\text{second difference}$
Alternating, grouped and greatest-integer sums
Alternating squares
$1^2-2^2+3^2-\dots-(2n)^2=-n(2n+1)$

Watch out for (3)

Sum first, then put in the limit
→ Σk, Σk² and Σk³ applied to a polynomial kth term
Check the fourth term
→ Finding the kth term from differences or from the sum
An odd count leaves one term unpaired
→ Alternating, grouped and greatest-integer sums

Test yourself on Sequences and Series

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Sequences and Series chapter test Drill all 34 questions