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If `sum_(i=1)^(n-1) ((""^(n)C_(i-1))/(""^(n)C_(i)+""^(n)C_(i-1)))^(3) = (36)/(13)`, , then n is equal toA. 10B. 11C. 13D. 12 |
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Answer» Correct Answer - d We have , `sum_(i-1)^(n-1) ((""^(n)C_(i-1))/(""^(n)C_(i)+""^(n)C_(i-1)))^(3) = (36)/(13)`, , `rArr sum_(i-1)^(n-1) ((""^(n)C_(i-1))/(""^(n+1)C_(i)))^(3) = `(36)/(13) ` [ `because ""^(n) C_(r-1) + ""^(n)C_(r) = ""^(n+1)C_(r)]` `rArr sum_(i-1)^(n-1) ((""^(n)C_(i-1))/((n+1)/(i)""^(n)C_(i-1)))^(3) = (36)/(13) ` `rArr sum_(i-1)^(n-1) ((i)/(n+1))^(3) = (36)/(13)` `rArr (1)/(n+3)^(3) {(n(n+1))/(2)}^(2) = (36)/(13)` `(n^(2))/(4 (n +1)^(3))= (36)/(13)` `rArr (n^(2))/ (n +1)=(144)/(13) rArr (n^(2))/(n+1) = (12^(2))/((12 +1))rArr n=12`. |
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