Question

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

• A. $a=\frac{1}{12}$
• B. $b=\frac{1}{16}$
• C. $d=\frac{1}{6}$
• D. $e=0$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $a=\frac{1}{12}$
C $d=\frac{1}{6}$
D $e=0$

${\sum }_{k=1}^n\left({\sum }_{m=1}^k{m}^2\right)=a{n}^4+b{n}^3+c{n}^2+dn+e$
$\Rightarrow {\sum }_{k=1}^n\left[\frac{k(k+1)(2k+1)}{6}\right]=a{n}^4+b{n}^3+c{n}^2+dn+e$
$\Rightarrow \frac{1}{6}\{2{\sum }_{k=1}^n{k}^3+3{\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^{n}k\}=a{n}^4+b{n}^3+c{n}^2+dn+e$
$\Rightarrow \frac{1}{6}\{\frac{k^2{(k+1)}^2}{2}+\frac{k(k+1)(2k+1)}{2}+\frac{k(k+1)}{2}\}=a{n}^4+b{n}^3+c{n}^2+dn+e$
$\Rightarrow \frac{n^4}{12}+\frac{n^3}{3}+\frac{{5n}^2}{12}+\frac{n}{6}=a{n}^4+b{n}^3+c{n}^2+dn+e$
Therefore, $a=\frac{1}{12}$; $b=\frac{1}{3}$; $c=\frac{5}{12}$; $d=\frac{1}{6}$; $e=0$
Ans: A,C,D