If the n^{th} term of a sequence is n(n + 1), | vedclass

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

Question

(A) The $n^{th}$ term is given by $a_n = n(n + 1) = n^2 + n$. To find the sum of $n$ terms,we calculate $S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The $n^{th}$ term is given by $a_n = n(n + 1) = n^2 + n$. To find the sum of $n$ terms,we calculate $S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + k)$. Using the standard summation formulas $\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}$,we get: $S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$. Factoring out $\frac{n(n + 1)}{2}$,we have: $S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1}{3} + 1 \right) = \frac{n(n + 1)}{2} \left( \frac{2n + 1 + 3}{3} \right)$. $S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 4}{3} \right) = \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{n(n + 1)(n + 2)}{3}$.

If the $n^{th}$ term of a sequence is $n(n + 1)$,then what is the sum of its $n$ terms?

Similar Questions

If the sum of the first $11$ terms of the series ${\left( {1\frac{4}{7}} \right)^2} + {\left( {1\frac{5}{7}} \right)^2} + {\left( {1\frac{6}{7}} \right)^2} + {2^2} + {\left( {2\frac{1}{7}} \right)^2} + \dots$ is $\frac{11}{7}\lambda$,then $\lambda$ is equal to:

The sum $1 + \frac{1}{2}(1^2+2^2) + \frac{1}{3}(1^2+2^2+3^2) + \dots$ up to $10$ terms is equal to:

If $\alpha, \beta$ are natural numbers such that $100^{\alpha} - 199\beta = (100)(100) + (99)(101) + (98)(102) + \ldots + (1)(199)$,then the slope of the line passing through $(\alpha, \beta)$ and the origin is:

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

The numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module