Find the sum of $1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . . + 9^{3}$

400

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425

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450

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475

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Solution

The correct option is C $425$
$1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . + 9^{3}$

$= 1^{3} + 2^{3} + 3^{3} + . . . + 9^{3} - 2 (2^{3} + 4^{3} + 6^{3} + 8^{3})$

$= [\frac{9 \times 10}{2}]^{2} - {2.2}^{3} [1^{3} + 2^{3} + 3^{3} + 4^{3}]$

$= (45)^{2} - 16 [\frac{4 \times 5}{2}]^{2}$

$= 2025 - 1600 = 425$

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

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