Question

The sum of first n terms of a sequence is $\frac{n^2(n+1{)}^2}{4}$. Find its nth term. Examine whether the sequence is an A.P. or a G.P.

Answer 1

Given $S_n=\frac{n^2(n+1{)}^2}{4}$
$\therefore S_{n-1}=\frac{(n-1{)}^2[(n-1)+1{]}^2}{4}=\frac{n^2(n-1{)}^2}{4}$
Now, $t_n=S_n-S_{n-1}$
$=\frac{n^2(n-1{)}^2}{4}-\frac{n^2(n-1{)}^2}{4}=\frac{n^2(n+1{)}^2-n^2(n-1{)}^2}{4}$
$=\frac{n^2\left[\right(n+1{)}^2-(n-1{)}^2]}{4}$
$=\frac{n^2[n^2+2n+1-(n^2-2n+1\left)\right]}{4}$
$=\frac{n^2[n^2+2n+1-n^2+2n-1]}{4}=\frac{n^2\left(4n\right)}{4}$
$\therefore t_n=n^3$
$\therefore$ Hence, $t_{n+1}=(n+1{)}^3$
Now, consider $t_{n+1}-t_n=(n+1{)}^3-n^3$
$=n^3+3n^2+3n+1-n^3=3n^2+3n+1$
$\ne$ constant.
$\therefore$ The given sequence is not in A.P.
Now, consider
$\frac{t_{n+1}}{t_n}=\frac{(n+1{)}^3}{n^3}\ne$ constant
$\therefore$ The sequence is not a G.P.
Hence, the given sequence is not an A.P. nor a G.P.