sumlimits_{i = 1}^n {sumlimits_{j = 1}^i {sum | vedclass

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

Question

(D) We evaluate the triple summation from the inside out:$\sum_{k=1}^j 1 = j$Next,$\sum_{j=1}^i j = \frac{i(i+1)}{2}$Finally,$\sum_{i=1}^n \frac{i(i+1

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) We evaluate the triple summation from the inside out:$\sum_{k=1}^j 1 = j$Next,$\sum_{j=1}^i j = \frac{i(i+1)}{2}$Finally,$\sum_{i=1}^n \frac{i(i+1)}{2} = \frac{1}{2} [\sum_{i=1}^n i^2 + \sum_{i=1}^n i]$Using the standard summation formulas $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum i = \frac{n(n+1)}{2}$:$= \frac{1}{2} [\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}]$$= \frac{n(n+1)}{4} [\frac{2n+1}{3} + 1]$$= \frac{n(n+1)}{4} [\frac{2n+4}{3}]$$= \frac{n(n+1)(n+2)}{6}$

$\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = \dots$

Similar Questions

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

Find the $9^{\text{th}}$ term in the sequence whose $n^{\text{th}}$ term is given by $a_{n} = (-1)^{n-1} n^{3}$.

What is the $99^{th}$ term of the series $2 + 7 + 14 + 23 + 34 + \dots$?

How many terms are there in the sequence $1, 3, 6, 10, 15, 21, \dots, 5050$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module