If $\sum_{r=1}^{n} r=55$, find $\sum_{r=1}^{n} r^{3} .$
$\sum_{r=1}^{n} r^{3}=1^{3}+2^{3}+3^{3}+\ldots+n^{3}$
$=\left[\frac{n(n+1)}{2}\right]^{2}$
$=\left[\sum_{r=1}^{n} r\right]^{2}$
$=[55]^{2}$
$=3025$
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