Question

If ${\sum }_{k=1}^n$ ${\sum }_{m=1}^k$ $m^2$ = a$n^4$ + b$n^3$ + c$n^2$ + dn + e. then b =

• A. 1 / 12
• B. 1 / 3
• C. 5 / 12
• D. 1 / 6
• E. 5 / 6

Answer 1

The correct option is B

1 / 3

${\sum }_{k=1}^n$ ${\sum }_{m=1}^k$ = a$n^4$ + b$n^3$ + c$n^2$ + dn + e -----------(1)

${\sum }_{k=1}^n$ ${\sum }_{m=1}^k$ $m^2$ = ${\sum }_{k=1}^n$ $\frac{k(k+1)(2k+1)}{6}$

= $\frac{1}{6}$ ${\sum }_{k=1}^n$ [$2k^3+3k^2+k$]

Here, ${\sum }_{k=1}^n$[$k^3$] = $\frac{n^2(n+1{)}^2}{4}$

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

${\sum }_{k=1}^n$[k] = $\frac{n(n+1)}{2}$

=$\frac{1}{6}$ $[2.\frac{n^2(n+1{)}^2}{4}+3.\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}]$

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

= $\frac{1}{12}$ n(n+1) [$n^2$ + n + 2n + 1 + 1]

= $\frac{1}{12}$ n [$n^3$ + 3$n^2$ + 2n + $n^2$ + 3n + 2]

= $\frac{n}{12}$ [$n^3$ + 4$n^2$ + 5n + 2]

= $\frac{n^4}{12}$ + $\frac{n^3}{3}$ + $\frac{5}{12}{n}^2$ + $\frac{n}{6}$

Compare it with equation a$n^4$ + b$n^3$ + c$n^2$ + dn + e.

We get,

a = $\frac{1}{12}$, b =$\frac{1}{3}$, c = $\frac{5}{12}$, d = $\frac{1}{6}$.