(C) The ${n^{th}}$ term is given by $T_n = \frac{\sum_{k=1}^{n} k^3}{\sum_{k=1}^{n} (2k - 1)}$.The numerator is the sum of cubes of the first $n$ natu

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(C) The ${n^{th}}$ term is given by $T_n = \frac{\sum_{k=1}^{n} k^3}{\sum_{k=1}^{n} (2k - 1)}$.The numerator is the sum of cubes of the first $n$ natural numbers: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$.The denominator is the sum of the first $n$ odd numbers,which is $n^2$: $\sum_{k=1}^{n} (2k - 1) = n^2$.Therefore,$T_n = \frac{n^2(n+1)^2 / 4}{n^2} = \frac{(n+1)^2}{4} = \frac{n^2 + 2n + 1}{4}$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Difficult

The ${n^{th}}$ term of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ is:

A
$n^2 + 2n + 1$
B
$\frac{n^2 + 2n + 1}{8}$
C
$\frac{n^2 + 2n + 1}{4}$
D
$\frac{n^2 - 2n + 1}{4}$

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Class 11

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Sequences and Series

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nth term of special series, Sum to n terms and Infinite number of terms

If $|x| < 1$,what is the sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots \infty$?

Medium

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If $\sum_{k=1}^{n} a_k = 6n^3$, then $\sum_{k=1}^{6} \left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to . . . . . . .

DifficultJEE MAIN 2026

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The sum of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$ up to $n$ terms is:

Medium

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If $0 < \theta, \phi < \frac{\pi}{2}$,$x = \sum_{n=0}^{\infty} \cos^{2n} \theta$,$y = \sum_{n=0}^{\infty} \sin^{2n} \phi$,and $z = \sum_{n=0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi$,then:

DifficultJEE MAIN 2021

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Find the $9^{\text{th}}$ term in the sequence whose $n^{\text{th}}$ term is given by $a_{n} = (-1)^{n-1} n^{3}$.

Easy

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The ${n^{th}}$ term of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ is:

Similar Questions

If $|x| < 1$,what is the sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots \infty$?

If $\sum_{k=1}^{n} a_k = 6n^3$, then $\sum_{k=1}^{6} \left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to . . . . . . .

The sum of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$ up to $n$ terms is:

If $0 < \theta, \phi < \frac{\pi}{2}$,$x = \sum_{n=0}^{\infty} \cos^{2n} \theta$,$y = \sum_{n=0}^{\infty} \sin^{2n} \phi$,and $z = \sum_{n=0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi$,then:

Find the $9^{\text{th}}$ term in the sequence whose $n^{\text{th}}$ term is given by $a_{n} = (-1)^{n-1} n^{3}$.

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