If the n^{th} term of a series is 3 + n(n - 1 | vedclass

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

Question

(B) Given the $n^{th}$ term $T_n = 3 + n(n - 1) = n^2 - n + 3$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 - k + 3)$.Using

Vedclass · Accepted Answer

$\frac{n^3 + 8n}{3}$

Vedclass · Answer

(B) Given the $n^{th}$ term $T_n = 3 + n(n - 1) = n^2 - n + 3$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 - k + 3)$.Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum k = \frac{n(n+1)}{2}$,and $\sum 1 = n$:$S_n = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + 3n$.$S_n = \frac{n(n+1)(2n+1) - 3n(n+1) + 18n}{6}$.$S_n = \frac{n[(n+1)(2n+1) - 3(n+1) + 18]}{6}$.$S_n = \frac{n[2n^2 + 3n + 1 - 3n - 3 + 18]}{6}$.$S_n = \frac{n[2n^2 + 16]}{6} = \frac{2n(n^2 + 8)}{6} = \frac{n(n^2 + 8)}{3} = \frac{n^3 + 8n}{3}$.

If the $n^{th}$ term of a series is $3 + n(n - 1)$,then the sum of $n$ terms of the series is

Similar Questions

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ up to the $11^{th}$ term is:

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms is:

The sum of the series $3.6 + 4.7 + 5.8 + \dots$ up to $(n - 2)$ terms is:

The value of $\frac{1 \times 2^2 + 2 \times 3^2 + \ldots + 100 \times 101^2}{1^2 \times 2 + 2^2 \times 3 + \ldots + 100^2 \times 101}$ is

$1 + (1 + a)x + (1 + a + a^2)x^2 + \dots \infty = \dots \, (0 < a, x < 1)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module