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

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Solution

= $1^{3} + 3^{3} + 5^{3} + 7^{3} + 9^{3} - (2^{3} + 4^{3} + 6^{3} + 8^{3})$
=Sum of cubes of first 5 odd number -Sum of cubes of first 4 even number
= $n^{2} (2 n^{2} - 1) - 2 n^{2} (n + 1)^{2}$
=1225–800
=425

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

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

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

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

(1^{3} + 2^{3} + 3^{3} + 4^{3})^{- 3 / 2}

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