The sum of n terms of the series 1^{3}+3^{3}+ | vedclass

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

Question

(A) The $n^{th}$ term of the series is $T_{n} = (2n-1)^{3}$.Expanding this, we get $T_{n} = 8n^{3} - 12n^{2} + 6n - 1$.The sum of $n$ terms is $S_{n}

Vedclass · Accepted Answer

$n^{2}(2n^{2}-1)$

Vedclass · Answer

(A) The $n^{th}$ term of the series is $T_{n} = (2n-1)^{3}$.Expanding this, we get $T_{n} = 8n^{3} - 12n^{2} + 6n - 1$.The sum of $n$ terms is $S_{n} = \sum_{k=1}^{n} T_{k} = \sum_{k=1}^{n} (8k^{3} - 12k^{2} + 6k - 1)$.Using standard summation formulas:$S_{n} = 8 \left[ \frac{n(n+1)}{2} \right]^{2} - 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] - n$.$S_{n} = 2n^{2}(n+1)^{2} - 2n(n+1)(2n+1) + 3n(n+1) - n$.$S_{n} = 2n^{2}(n^{2}+2n+1) - 2n(2n^{2}+3n+1) + 3n^{2} + 3n - n$.$S_{n} = 2n^{4} + 4n^{3} + 2n^{2} - 4n^{3} - 6n^{2} - 2n + 3n^{2} + 2n$.$S_{n} = 2n^{4} - n^{2} = n^{2}(2n^{2}-1)$.

The sum of $n$ terms of the series $1^{3}+3^{3}+5^{3}+7^{3}+\ldots$ is

Similar Questions

Let $S_k = \frac{1 + 2 + 3 + .... + k}{k}$. If $S_1^2 + S_2^2 + ....... + S_{10}^2 = \frac{5}{12}A$,then $A$ is equal to:

Find the sum to $n$ terms of the series $3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + \dots$

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

If $a, x$ are real numbers and $|a| < 1, |x| < 1$, then $1 + (1+a)x + (1+a+a^2)x^2 + \dots \infty$ is equal to

Let $a_n$ be a sequence such that $a_1 = 5$ and $a_{n+1} = a_n + (n - 2)$ for all $n \in N$. Then $a_{51}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module