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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
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 ---- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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
----
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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