The sum of n terms of the series whose n^{th} | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{n(n + 1)(n + 2)}{3}$

Vedclass · Answer

(A) Given the $n^{th}$ term is $T_n = n(n + 1) = n^2 + n$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.Using the standard summation formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$,we get:$S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$.Factor out $\frac{n(n + 1)}{2}$:$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1}{3} + 1 \right)$.$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1 + 3}{3} \right) = \frac{n(n + 1)(2n + 4)}{6}$.$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{n(n + 1)(n + 2)}{3}$.

The sum of $n$ terms of the series whose $n^{th}$ term is $n(n + 1)$ is equal to

Similar Questions

The sum to infinite terms of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ is

The sum of the infinite series $\frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + ..........\infty$ is equal to :-

If $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$,then

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ are in $H.P.$,then ${a_1}{a_2} + {a_2}{a_3} + \dots + {a_{n-1}}{a_n}$ is equal to:

If the sum of $n$ terms of an $A.P.$ is $3n^2 + 5n$ and $T_m = 164$,then $m = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module