If sum_{i = 1}^n sum_{j = 1}^i sum_{k = 1}^j | vedclass

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

Question

(B) Given the expression: $\sum_{i = 1}^n \sum_{j = 1}^i \sum_{k = 1}^j 1 = 560$.First,evaluate the innermost sum: $\sum_{k = 1}^j 1 = j$.Now,the expr

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) Given the expression: $\sum_{i = 1}^n \sum_{j = 1}^i \sum_{k = 1}^j 1 = 560$.First,evaluate the innermost sum: $\sum_{k = 1}^j 1 = j$.Now,the expression becomes: $\sum_{i = 1}^n \sum_{j = 1}^i j = 560$.Next,evaluate the middle sum: $\sum_{j = 1}^i j = \frac{i(i + 1)}{2}$.Now,the expression becomes: $\sum_{i = 1}^n \frac{i(i + 1)}{2} = 560$.This can be written as: $\frac{1}{2} [\sum_{i = 1}^n i^2 + \sum_{i = 1}^n i] = 560$.Using the summation formulas $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum i = \frac{n(n+1)}{2}$:$\frac{1}{2} [\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}] = 560$.Factor out $\frac{n(n+1)}{2}$:$\frac{1}{2} \cdot \frac{n(n+1)}{2} [\frac{2n+1}{3} + 1] = 560$.$\frac{n(n+1)}{4} [\frac{2n+4}{3}] = 560$.$\frac{n(n+1) \cdot 2(n+2)}{12} = 560$.$\frac{n(n+1)(n+2)}{6} = 560$.$n(n+1)(n+2) = 560 	imes 6 = 3360$.Since $14 	imes 15 	imes 16 = 3360$,we have $n = 14$.

If $\sum_{i = 1}^n \sum_{j = 1}^i \sum_{k = 1}^j 1 = 560$,then the value of $n$ is:

Similar Questions

If $x = \sum_{n = 0}^\infty a^n$,$y = \sum_{n = 0}^\infty b^n$,and $z = \sum_{n = 0}^\infty (ab)^n$,where $a, b < 1$,then:

The value of $\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to

$\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{99 \times 100}\right]$ is equal to

$0.5737373...... = $

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module