The value of 1^3 - 2^3 + 3^3 - dots + 15^3 is | vedclass

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(B) Let the series be $S = 1^3 - 2^3 + 3^3 - 4^3 + \dots + 15^3$.We can group the terms as $S = (1^3 + 3^3 + \dots + 15^3) - (2^3 + 4^3 + \dots + 14^3

Vedclass · Accepted Answer

$1856$

Vedclass · Answer

(B) Let the series be $S = 1^3 - 2^3 + 3^3 - 4^3 + \dots + 15^3$.We can group the terms as $S = (1^3 + 3^3 + \dots + 15^3) - (2^3 + 4^3 + \dots + 14^3)$.The sum of odd cubes is $\sum_{k=1}^8 (2k-1)^3 = \sum_{k=1}^8 (8k^3 - 12k^2 + 6k - 1)$.Using standard summation formulas:$\sum_{k=1}^8 k^3 = [8(9)/2]^2 = 36^2 = 1296$.$\sum_{k=1}^8 k^2 = 8(9)(17)/6 = 204$.$\sum_{k=1}^8 k = 8(9)/2 = 36$.Sum of odd cubes $= 8(1296) - 12(204) + 6(36) - 8 = 10368 - 2448 + 216 - 8 = 8128$.The sum of even cubes is $\sum_{k=1}^7 (2k)^3 = 8 \sum_{k=1}^7 k^3 = 8 [7(8)/2]^2 = 8(28^2) = 8(784) = 6272$.Therefore, $S = 8128 - 6272 = 1856$.

The value of $1^3 - 2^3 + 3^3 - \dots + 15^3$ is:

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