If `(1^2-t_1)+(2^2-t_2)+---+(n^2-t_n)=(n(n^2-1))/3` , then `t_n` is equal to a. `n^2` b. `2n` c. `n^2-2n` d. none of these

Options

A. `n^2`
B. 2n
C. `n^2-2n`
D. none of these

Answer

D

Explanation

`(1^(2)-t_(1))+(2^(2)-t^(2))+…+(n^(2)-t_(n))=1/3n(n^(2)-1)`
`rArr1^(2)+2^(2)+3^(2)+…+n^(2)-{t_(1)+t_(2)+…+t_(n)}=1/3n(n^(2)-1)`
`rArrS_(n)=(n(n+1))/6[2n+1-2(n-1)]`
`=(n(n+1))/6[2n+1-2n+2]`
`=(n(n+1))/2`
`=S_(n-1)=(n(n-1))/2`
`rArrt_(n)=S_(n)-S_(n-1)=n`