For any integer n geq 1,the sum sum_{k=1}^n k | vedclass

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

Question

(C) We need to evaluate the sum $S = \sum_{k=1}^n k(k+2)$.Expanding the term inside the summation,we get $k^2 + 2k$.Thus,$S = \sum_{k=1}^n (k^2 + 2k)

Vedclass · Accepted Answer

$\frac{n(n+1)(2n+7)}{6}$

Vedclass · Answer

(C) We need to evaluate the sum $S = \sum_{k=1}^n k(k+2)$.Expanding the term inside the summation,we get $k^2 + 2k$.Thus,$S = \sum_{k=1}^n (k^2 + 2k) = \sum_{k=1}^n k^2 + 2 \sum_{k=1}^n k$.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 have:$S = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.$S = \frac{n(n+1)(2n+1)}{6} + n(n+1)$.Factoring out $\frac{n(n+1)}{6}$,we get:$S = \frac{n(n+1)}{6} [ (2n+1) + 6 ]$.$S = \frac{n(n+1)(2n+7)}{6}$.

For any integer $n \geq 1$,the sum $\sum_{k=1}^n k(k+2)$ is equal to

Similar Questions

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$

If $f(1)=3$, and $f(n+1)-f(n)=3(4^n-1)$, then for all $n \in N$, $f(n)=$

If $2^3+4^3+6^3+\ldots+(2n)^3 = h n^2(n+1)^2$,then $h$ is equal to

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{n}{n+1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module