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If $S_n=1^3+2^3+\ldots+n^3$ and $T_n=1+2+\ldots+n$, then
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Verified Answer
The correct answer is:
$S_n=T_n^2$
Given, $S_n=1^3+2^3+\ldots+n^3=\Sigma n^3$
$\begin{array}{rlrl} & \text { and } & T_n & =1+2+\ldots+n=\Sigma n \\ & & S_n & =\Sigma n^3=\left[\frac{n(n+1)}{2}\right]^2 \\ & & =\{\Sigma(n)\}^2=T_n^2 \\ & \therefore & S_n & =T_n^2\end{array}$
$\begin{array}{rlrl} & \text { and } & T_n & =1+2+\ldots+n=\Sigma n \\ & & S_n & =\Sigma n^3=\left[\frac{n(n+1)}{2}\right]^2 \\ & & =\{\Sigma(n)\}^2=T_n^2 \\ & \therefore & S_n & =T_n^2\end{array}$
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