COMPASSLearning

Talk to us Register Login

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Sequences and Series

The sum of the following series

1 + 6 + \frac{9 ( 1 ^{2} + 2 ^{2} + 3 ^{2} )}{7} + \frac{12 ( 1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} )}{9} + \frac{15 ( 1 ^{2} + 2 ^{2} + \dots . + 5 ^{2} )}{11} + \dots

up to 15 terms, is:

(A)7820✓ Correct
(B)7830
(C)7520
(D)7510

Answer: (A)

Solution

The general term is

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

. Using

1^{2} + \dots + n^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

, the

(2 n + 1)

cancels and

T_{n} = \frac{3 n}{2 n + 1} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{n ^{2} ( n + 1 )}{2}

, which checks out as

T_{1} = 1

and

T_{2} = 6

. Summing to

15

terms,

S_{15} = \frac{1}{2} \sum_{n = 1}^{15} (n^{3} + n^{2}) = \frac{1}{2} [(\frac{15 \cdot 16}{2})^{2} + \frac{15 \cdot 16 \cdot 31}{6}] = \frac{1}{2} [14400 + 1240] = 7820

.

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Sequences and Series →

More JEE Main questions on Sequences and Series

Let $a_{1}, a_{2}, a_{3}, \dots$ . be a G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ .

Then $a_{3} + a_{5} + a_{7}$ is equal to :
JEE Main 2026View solution →
Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to
JEE Main 2026View solution →
The positive integer n, for which the solutions of the equation
$x (x + 2) + (x + 2) (x + 4) + \dots + (x + 2 n - 2) (x + 2 n) = \frac{8 n}{3}$

are two consecutive even integers, is :-
JEE Main 2026View solution →
Let $a_{1}, \frac{a _{2}}{2}, \frac{a _{3}}{2 ^{2}}, \dots, \frac{a _{10}}{2 ^{9}}$ be a G.P. of common ratio $\frac{1}{2}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to :
JEE Main 2026View solution →
If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
JEE Main 2026View solution →
Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ , then $9 (a^{2} + b^{2} + c^{2})$ is equa …
JEE Main 2026View solution →