InterviewSolution
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InterviewSolution
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For and odd integer `n ge 1, n^(3) - (n - 1)^(3) ` + …… `+ (- 1)^(n-1) 1^(3)` |
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Answer» Since, n is an odd interger, `(-1)^(n-1)=1` `and n-1,n-3,n-5` etc., are even intergers, then `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)]` `=sumn^(3)-2xx2^(3)[((n-1)/(2))^(3)+((n-3)/(2))^(3)+ . . .+1^(3)]` `[beausen-1,n-3` . . ., are even intergers] `=sumn^(3)-16[sum((n-1)/(2))^(3)]` `=[(n(n+1)/(2)]^(2)-16[(1)/(2)((n-1)/(2))((n-1)/(2)+1)]^(2)` `=(1)/(4)(n+1)^(2)-(16(n-1)^(2)(n+1)^(2))/(4xx4xx4)` `=(1)/(4)(n+1)^(2)[n^(2)-(n-1)^(2)]=(1)/(4)(n+1)^(2)(2n-1)` |
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