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 $t_n$.

• A. n
• B. $2$n
• C. $3$n
• D. $4$n

Answer 1

The correct option is A n

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

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

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

$=\frac{2n^3+3n^2+n}{6}-(\frac{2n^3}{6}-2n)$

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

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

$\therefore t_1=1$

$\therefore t_2=3-1=2$

$\therefore t_3=6-3=3$

$\therefore t_n=n$