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

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(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:

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