InterviewSolution
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InterviewSolution
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Thesum of first 9 terms of the series `(1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+2+3)+. . . . .`is(1)71 (2) 96 (3) 142 (4) 192 |
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Answer» `T_n = (1^3 + 2^3 + 3^3 + ...... + n^3)/(1 + 3 + 5 + 7 + ...... + (2n-1))` `= ((n(n+1))/2)^2/(n/2[1 + (2n-1)])` `= n^2(n+1)^2/(4 xx n^2)` `T_n = (n+1)^2/4` `sum_(x=1)^9 T_n = sum_(n=1)^9 (n+1)^2/4` `= 1/4 sum_(n=1)^9 (n+1)^2` = `1/4 ( 2^2 + 3^2 + 4^2 + ...+ 10^2)` `= 1/4 ( 1^2 + 2^2 + 3^2 + .... + 10^2 - 1^2)` `= 1/4[(10(10+1)(2 xx 10 + 1))/6 - 1]` `= 1/4[(10 xx 11 xx 27)/6- 1]` `= 1/4 xx 384= 96` option 1 is correct |
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