The {n^{th}} term of the series frac{1^3}{1} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{n^2 + 2n + 1}{4}$

Vedclass · Answer

(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,$T_n = \frac{n^2(n+1)^2 / 4}{n^2} = \frac{(n+1)^2}{4} = \frac{n^2 + 2n + 1}{4}$.

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

The sum to $n$ terms of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ is

The sum of $n$ terms of the following series $1 + (1 + x) + (1 + x + x^2) + \dots$ will be

The sum of the series: $(2)^2 + 2(4)^2 + 3(6)^2 + \dots$ up to $10$ terms is

Find the sum of the series $2 + 4 + 7 + 11 + 16 + \dots$ up to $n$ terms.

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module