The sum of $1^{3}$ + $2^{3}$ + $3^{3}$ + $4^{3}$ + ........... + $15^{3}$ is

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Solution

Sum of cubes of 'n' natural number

= $\frac{n^{2} (n + 1)^{2}}{4}$ = $\frac{15^{2} (16)^{2}}{4}$ = 14,400.

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Similar questions

1^{3} + 2^{3} + 3^{3} + 4^{3} + \dots + 15^{3}

1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . . + 9^{3}

The sum of the series 1³ - 2³+ 3³- 4³+ ....... + 9³ is:

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

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