Question

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

• A. $\frac{3n^2-2n+1}{2}$
• B. $n-1$
• C. $n+1$
• D. $n$

Answer 1

The correct option is A $\frac{3n^2-2n+1}{2}$

$1^2+2^2+3^2+....n^2-(t_1+t_2+....+t_n)=\left(\frac{1}{3}\right)(n^2-1)\left(\frac{n(n+1)(2n+1)}{6}\right)-\left(\frac{1}{3}\right)(n^2-1)=\sum tn\left(\frac{n+1}{3}\right)\left[\right(\frac{n(2n+1)}{2})-(n-1\left)\right]=\sum tn\left(\frac{n+1}{3}\right)\left(\frac{n(2n+1-2(n-1\left)\right)}{2}\right)=\sum tn\left(\frac{n+1}{3}\right)\left(\frac{2n^2+n-2n+2}{2}\right)=\sum tn\left(\frac{n+1}{3}\right)\left(\frac{2n^2-n+2}{2}\right)=\sum tn\therefore s_n=\left(\frac{(n+1)(2n^2-n+2)}{6}\right)t_n=s_n-s_{n-1}=\left(\frac{(n+1)(2n^2-n+2)}{6}\right)-\left(\frac{n\left[2\right(n-1{)}^2-(n-1)+2]}{6}\right)=\left(\frac{1}{6}\right)\left[\right(n+1\left)\right(2n^2-n+2\left)\right]-n(2n^2-4n+2-n+1+2)=\left(\frac{1}{6}\right)[2n^3-n^2+2n+2n^2-n+2-2n^3+5n^2-5n]\left(\frac{1}{6}\right)(6n^2-4n+2)=\left(\frac{3n^2-2n+1}{3}\right)$