$1^{3} + 2^{3} + 3^{3} + . . . + 10^{3} =$ ?

3025

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3500

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3250

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4250

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Solution

The correct option is A 3025
The sum of cubes of first $n$ natural numbers is given by
$S_{n} = [\frac{n (n + 1)}{2}]^{2} .$
$\Rightarrow 1^{3} + 2^{3} + 3^{3} + . . . + 10^{3} = [\frac{10 \times 11}{2}]^{2} = [55]^{2} = 3025$

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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} + \frac{1^{3} + 2^{3}}{2} + \frac{1^{3} + 2^{3} + 3^{3}}{3} + . . .

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

=

k

1^{3} + 2^{3} + 3^{3} + . . . + k^{3} = 6084

1^{3} + 2^{3} + 3^{3} + . . . . + k^{3} = 2025

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