Find the sum of the first n terms of the seri | vedclass

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

Question

(A) The given series is $3+7+13+21+31+\ldots$Let $a_n$ be the $n^{th}$ term of the series.The differences between consecutive terms are $4, 6, 8, 10,

Vedclass · Accepted Answer

$\frac{n}{3}(n^2+3n+5)$

Vedclass · Answer

(A) The given series is $3+7+13+21+31+\ldots$Let $a_n$ be the $n^{th}$ term of the series.The differences between consecutive terms are $4, 6, 8, 10, \ldots$,which form an arithmetic progression.Thus,$a_n = a_1 + \sum_{k=1}^{n-1} (4 + (k-1)2) = 3 + \sum_{k=1}^{n-1} (2k+2) = 3 + 2\frac{(n-1)n}{2} + 2(n-1) = 3 + n^2 - n + 2n - 2 = n^2 + n + 1$.Now,the sum of the first $n$ terms is $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + k + 1)$.$S_n = \sum_{k=1}^n k^2 + \sum_{k=1}^n k + \sum_{k=1}^n 1$.$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n$.$S_n = \frac{n}{6} [ (n+1)(2n+1) + 3(n+1) + 6 ]$.$S_n = \frac{n}{6} [ 2n^2 + 3n + 1 + 3n + 3 + 6 ]$.$S_n = \frac{n}{6} [ 2n^2 + 6n + 10 ]$.$S_n = \frac{n}{3} [ n^2 + 3n + 5 ]$.

Find the sum of the first $n$ terms of the series: $3+7+13+21+31+\ldots$

Similar Questions

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots$ is equal to

The $n^{th}$ term of the following series $(1 \times 3) + (3 \times 5) + (5 \times 7) + (7 \times 9) + \dots$ will be

The sum to $n$ terms of $(2n - 1) + 2(2n - 3) + 3(2n - 5) + \dots$ is

$11^2 + 12^2 + 13^2 + \dots + 20^2 = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module