Question

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

• A. $a+c=b+d$
• B. $e=0$
• C. $a,b-\frac{2}{3},c-1$ are in A.P.
• D. $\frac{c}{a}$ is an integers

Answer 1

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

The correct options are
A $e=0$
B $a,b-\frac{2}{3},c-1$ are in A.P.
C $\frac{c}{a}$ is an integers
D $a+c=b+d$

$\sum _{r=1}^{n}r(r+1)(2r+3)=\sum _{r=1}^n({2r}^3+{3r}^2+{2r}^2+3r)=\sum _{r=1}^n({2r}^3+{5r}^2+3r)$
$=2{\left(\frac{n(n+1)}{2}\right)}^2+5\left(\frac{n(n+1)(2n+1)}{6}\right)+3\left(\frac{n(n+1)}{2}\right)$
$=\frac{1}{2}\left(n^2(n^2+1+2n)\right)+\frac{5}{6}\left(n({2n}^2+3n+1)\right)+\frac{3}{2}(n^2+n)$
$=\frac{1}{2}(n^4+{2n}^3+n^2)+\frac{5}{6}({2n}^3+{3n}^2+n)+\frac{3}{2}(n^2+n)$
$=\frac{1}{2}{n}^4+\frac{16}{6}{n}^3+\frac{9}{2}{n}^2+\frac{14}{6}n$