Find the value of n?1, n^3 � (n-1)^3 + ���.+ (-1)^n-1 1^3 =����for any odd integer.

Since n is an odd integer, (-1)^n-1 = 1 and n – 1, n – 3, n – 5, … are even integers.

We have

n^3 + (n – 1)^3 + (n – 2)^3 – (n – 3)^3 + …. + (-1)^n – 1 1^3

= n^3 + (n – 1)^3 + ( n – 2)^3 + ….. + 1^3 – 2[(n – 1)^3 + (n – 3)^3 + …… 2^3 ]

=[n^3 + (n – 1)^3 +( n – 2)^3 +……. + 1^3]

-2 * 2^3 [(n – 1/2)^3 + (n – 3/2)^3 + ….. + 1^3 ]

[ ∵ n – 1, n – 3 …………………….. are even integers]

Here the first square bracket contain the sum of cubes of Ist n natural numbers. Whereas the second square bracket contains the sum of the cubes of natural numbers from l to

(n – 1 /2), where n – 1, n – 3, ……………… are even integers. Using the formula for sum of cubes of lst n natural numbers we get the summation

= [ n (n + 1)/2]^2 – 16 [ { 1/2 ( n – 1/2)( n – 1/2 + 1)]^2

= 1/4n2 (n +1)^2 – 16 (n – 1)^2 ( n + 1)^2 /16 * 4

= 1/4 ( n + 1)^2 [n2 – ( n – 1)^2

= 1/4 ( n + 1)^2 (2n – 1)

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