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If n is an odd integergreater than or equal to 1 , then show that the value of
n3 - (n-1 ) 3 + (n-2)3 - ............+ (-1) n-1 . 1 3 is
(n+1)2 (2n-1) / 4
n3 - (n-1 )3 + (n-2)3 - ............+ (-1)n-1.13
2* [ n3 + (n-2)3 + (n-4)3 ............+ 23 ] - [ n3 + (n-1 )3 + (n-2)3 + ............+ 23 + 13 ]
2*23* [ ((n-1)/2)3 + ((n-4)/2)3 ............+ 13 ] - [ n3 + (n-1 )3 + (n-2)3 + ............+ 23 + 13 ]
sum of cubic terms of first n natural no's is: n2(n+1)2/4
so, 16*(n-1)2(n+1)2/64 - n2(n+1)2/4
(2n-1)(n+1)2 / 4
hence solved
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