If $1^{3} + 2^{3} + . . . . . + m^{3} = 3025$ , find m

12

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10

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9

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6

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Solution

The correct option is B $10$
Sum of cubes of 1st $m$ natural numbers is given by: $\frac{m^{2} (m + 1)^{2}}{4}$
$3025 = \frac{m^{2} (m + 1)^{2}}{4}$
$12100 = m^{2} (m + 1)^{2}$
Taking square roots on both sides we get,
$110 = m (m + 1)$
$m^{2} + m - 110 = 0$
$(m - 10) (m + 11) = 0$
$m = 10$ or $m = - 11$
Since $m$ cannot be negative, $m = 10$ .

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