If $1 + 2 + 3 + . . . . + n = 120$ , find $1^{3} + 2^{3} + 3^{3} + . . . n^{3}$

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Solution

Given $1 + 2 + 3 + . . . + n = 120$ i.e. $\frac{n (n + 1)}{2} = 120$
Therefore, $1^{3} + 2^{3} + . . . . + n^{3} = {(\frac{n (n + 1)}{2})}^{2} = 120^{2} = 14400$

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Similar questions

1^{3} + 2^{3} + 3^{3} + . . . . . . + n^{3} = {(\frac{n (n + 1)}{2})}^{2}

1^{3} + 2^{3} + 3^{3} + . . . . . . + 21^{3}

1^{3} + 2^{3} + 3^{3} + . . . . . . . . . + n^{3}

1^{3} + 2^{3} + 3^{3} + . . . . + n^{3} = 36100

1 + 2 + 3 + . . . . + n

S_{n} = \frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + \frac{1 + 2 + 3}{1^{3} + 2^{3} + 3^{3}} + . . . . . . + \frac{1 + 2 + . . . . . + n}{1^{3} + 2^{3} + . . . . + n^{3}}

100 S_{n} = n

n

S_{n} = \frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + . . . + \frac{1 + 2 + 3 + . . . + n}{1^{3} + 2^{3} + 3^{3} . . . + n^{3}}, n = 1, 2, 3 . . .

S_{n}

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