JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

n

is an odd integer greater than or equal to 1, then the value of

n^3-(n-1)^3+(n-2)^3-\ldots .+(-1)^{n-1} 1^3

, is

View answer and explanation

Correct option: A

\frac{(n+1)^2(2 n-1)}{4}

Option A (

\frac{(n+1)^2(2 n-1)}{4}

) is correct because it represents the value of the given alternating sum of cubes when n is an odd integer.

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