Q46 / 160 · AP EAMCET 2007 · Mathematics

If $S_n=1^3+2^3+\ldots+n^3$ and $T_n=1+2+\ldots+n$, then

  • A
    $S_n=T_{n^3}$
  • B
    $S_n=T_{n^2}$
  • Correct answer
    C
    $S_n=T_n^2$
  • D
    $S_n=T_n^3$

Solution

Given, $S_n=1^3+2^3+\ldots+n^3=\Sigma n^3$ $\begin{array}{rlrl} & \text { and } & T_n & =1+2+\ldots+n=\Sigma n \\ & & S_n & =\Sigma n^3=\left[\frac{n(n+1)}{2}\right]^2 \\ & & =\{\Sigma(n)\}^2=T_n^2 \\ & \therefore & S_n & =T_n^2\end{array}$

Hence, the correct answer is option: 'C'.

Relevant Questions