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Q.
If $\left(1^{2}-t_{1}\right)+\left(2^{2}-t_{2}\right)+\ldots .+\left(n^{2}-t_{n}\right)=\frac{n\left(n^{2}-1\right)}{3}$, then $t_{n}$ is equal to
Sequences and Series
Solution:
$\left(1^{2}-t_{1}\right)+\left(2^{2}-t_{2}\right)+\ldots .+\left(n^{2}-t_{n}\right)=\frac{1}{3} n\left(n^{2}-1\right)$
$\Rightarrow 1^{2}+2^{2}+3^{2}+\ldots .+n^{2}-\left\{ t _{1}+ t _{2}+\ldots .+ t _{ n }\right\}=\frac{1}{3} n \left( n ^{2}-1\right)$
$\Rightarrow \frac{ n ( n +1)(2 n +1)}{6}- S _{ n }=\frac{1}{3} n \left( n ^{2}-1\right)$
$\Rightarrow S _{ n }=\frac{ n ( n +1)}{6}[2 n +1-2( n -1)]$
$=\frac{ n ( n +1)}{6}[2 n +1-2 n +2]$
$=\frac{ n ( n +1)}{2}$
$\Rightarrow S _{ n -1}=\frac{ n ( n -1)}{2}$
$\Rightarrow T _{ n }= S _{ n }- S _{ n -1}= n$