For any integer n geq 1,sum_{K=1}^n K(K+2) = | vedclass

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

Question

(B) Consider the sum: $\sum_{k=1}^n k(k+2) = \sum_{k=1}^n (k^2 + 2k)$.Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Consider the sum: $\sum_{k=1}^n k(k+2) = \sum_{k=1}^n (k^2 + 2k)$.Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$:$\sum_{k=1}^n k^2 + 2 \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.Factor out $\frac{n(n+1)}{6}$:$= \frac{n(n+1)}{6} [ (2n+1) + 6 ]$.$= \frac{n(n+1)(2n+7)}{6}$.

For any integer $n \geq 1$,$\sum_{K=1}^n K(K+2) =$

Similar Questions

The $n^{th}$ term of the series $\frac{1}{1} + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$ is:

What is the $6^{th}$ term of the sequence $2, 1\frac{3}{4}, 1\frac{5}{9}, \dots$?

The sum of the series $6 + 66 + 666 + \dots$ up to $n$ terms is

$\sum_{n=1}^5 n(n^2+n+1) = $

The value of $5^{1/2} \cdot 5^{1/4} \cdot 5^{1/8} \cdots \infty$ is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module