InterviewSolution
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InterviewSolution
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Let `sum_(r=1)^(n) r^(6)=f(n)," then "sum_(n=1)^(n) (2r-1)^(6)` is equal toA. `f(n)-64f((n+1)/(2))` n is oddB. `f(n)-64f((n-1)/(2))` n is oddC. `f(n)-64f((n)/(2))`, n is evenD. none of these |
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Answer» Correct Answer - D we have, `underset(r=1)overset(n)sum(2r-1)^(6)` `=1^(6)+3^(6)+5^(6)+ . . .+(2n-1)^(6)` `={1^(6)+2^(6)+3^(6)+ . . . +(2n)^(6)}-{2^(6)+4^(6)+6^(6)+ . . . +(2n)^(6)}` `=f(2n)-2^(6)f(n)=f(2n)-64f(n)` |
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