Question

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

• A. $a+c=b+d$
• B. $d>a+c$
• C. $a+b=c$
• D. $b=2d+e$

Answer 1

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

The correct options are
A $a+c=b+d$
C $a+b=c$
D $b=2d+e$

$\sum _{k=1}^n\sum _{m=1}^k{m}^2=\sum _{k=1}^n\frac{k(k+1)(2k+1)}{6}=\frac{1}{6}\sum _{k=1}^n(2k^3+3k^2+k)=\frac{1}{3}{\left(\frac{n(n+1)}{2}\right)}^2+\frac{1}{2}\frac{\left(n\right)(n+1)(2n+1)}{6}+\frac{1}{6}\frac{n(n+1)}{2}=\frac{n^4+4n^3+5n^2+2n}{12}a=\text{coefficient of}{n}^4=\frac{1}{12}b=\text{coefficient of}{n}^3=\frac{1}{3}c=\text{coefficient of}{n}^2=\frac{5}{12}d=\text{coefficient of}n=\frac{1}{6}e=0$