Question

If $(1^2-t_1)+(2^2-t_2)+.....+(n^2-t_n)=\frac{1}{3}n(n^2-1)$, then $t_n$ is

• A. $\frac{n}{2}$
• B. $n-1$
• C. $n+1$
• D. $n$

Answer 1

The correct option is D $n$

The given series is $(1^2-t_1)+(2^2-t_2)+.........+(n^2-t_n)$

$=(1^2+2^2+3^2+.......+n^2)-$ $(t_1+t_2+t_3.......+t_n)$

$\frac{n(n+1)(2n+1)}{6}-$ $(t_1+t_2+t_3.......+t_n)$ $=\frac{1}{3}n(n^2-1)$

$\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$-$(t_1+t_2+t_3.......+t_n)$ $=\frac{1}{3}(n^3-n)$

$(t_1+t_2+t_3.......+t_n)$ $=\frac{n^2}{2}+\frac{n}{6}+\frac{n}{3}$

$(t_1+t_2+t_3.......+t_n)$ $=\frac{n^2}{2}+\frac{n}{2}$

$(t_1+t_2+t_3.......+t_n)$ $=\frac{n(n+1)}{2}$

So from this we can see that this is the sum first n natural numbers

$t_n=n$