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

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(B) We are given the triple summation: $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = 560$. First,evaluate the innermost

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) We are given the triple summation: $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j 1 } } = 560$. First,evaluate the innermost sum: $\sum\limits_{k = 1}^j 1 = j$. Now,the expression becomes: $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i j } = 560$. Next,evaluate the middle sum: $\sum\limits_{j = 1}^i j = \frac{i(i+1)}{2}$. Now,the expression becomes: $\sum\limits_{i = 1}^n \frac{i(i+1)}{2} = 560$. This can be written as: $\frac{1}{2} \left[ \sum\limits_{i=1}^n i^2 + \sum\limits_{i=1}^n i ight] = 560$. Using the standard summation formulas: $\frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ight] = 560$. Factoring out $\frac{n(n+1)}{2}$: $\frac{1}{2} \cdot \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 ight] = 560$. $\frac{n(n+1)}{4} \left[ \frac{2n+4}{3} ight] = 560$. $\frac{n(n+1) \cdot 2(n+2)}{12} = 560$. $\frac{n(n+1)(n+2)}{6} = 560$. $n(n+1)(n+2) = 560 	imes 6 = 3360$. We look for three consecutive integers whose product is $3360$. Since $14 	imes 15 	imes 16 = 3360$,we have $n = 14$.

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

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