Question

Find the value of expression ${\sum }_{i=1}^n{\sum }_{j=1}^i{\sum }_{k=1}^{j}6.$

• A. n(n+1)(n+3)
• B. n(n+1)(n+2)
• C. n(n+2)(n+3)
• D.

Answer 1

The correct option is B

n(n+1)(n+2)

${\sum }_{i=1}^n{\sum }_{j=1}^i{\sum }_{k=1}^{j}6.$

Let's calculate the summation part one by one from right hand side

= ${\sum }_{i=1}^n{\sum }_{j=1}^{i}6j.$ {Where ${\sum }_{k=1}^j$6=6j}

= 6${\sum }_{i=1}^n$ $\frac{i(i+1)}{2}$ {Where ${\sum }_{k=1}^j$ 6j = 6 $\times$ sum of first n natural number}

= 3[${\sum }_{i=1}^n{i}^2+{\sum }_{i=1}^{n}i$] where ${\sum }_{i=1}^n{i}^2$ = $\frac{n(n+1)(2n+1)}{6}$

= 3[$\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$]

= $\frac{3}{2}$ n(n+1) [$\frac{2n+1}{3}$ + 1]

= $\frac{3}{2}$ n(n+1) $\frac{2n+4}{3}$ = n(n+1)(n+2)