Question

If $(1^2-t_1)+(2^2-t_2)+.....+(n^2-t_n)=\frac{1}{3}\left[n\right(n^2-1\left)\right]$ then find the value of $t_n$.

Answer 1

Given: $(1^2-t_1)+(2^2-t_2)+.....+(n^2-t_n)=\frac{1}{3}\left[n\right(n^2-1\left)\right]$
$⟹t_1+t_2+...t_n=1^2+2^2+...n^2-\frac{1}{3}\left[n\right(n^2-1\left)\right]$
$⟹t_1+t_2+...t_n=\frac{n(n+1)(2n+1)}{6}-\frac{1}{3}\left[n\right(n^2-1\left)\right]$
$⟹t_1+t_2+...t_n=\frac{n}{6}\left[\right(n+1\left)\right(2n+1)-2(n^2-1\left)\right]$
$⟹t_1+t_2+...t_n=\frac{n}{6}[2n^2+3n+1-2n^2+2]$
$⟹t_1+t_2+...t_n=\frac{n}{6}[3n+3]$
$⟹t_1+t_2+...t_n=\frac{1}{2}[n^2+n]$ (1)
As the above is true for all $n$, replacing $n$ with $n-1$ gives:
$⟹t_1+t_2+...t_{n-1}=\frac{1}{2}\left[\right(n-1{)}^2+n-1]$
$⟹t_1+t_2+...t_{n-1}=\frac{1}{2}[n^2-n]$ (2)
Subtracting (2) from (1) gives:
$t_n=\frac{1}{2}\left[2n\right]=n$