InterviewSolution
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InterviewSolution
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Find the sum of series upto n terms `((2n+1)/(2n-1))+3((2n+1)/(2n-1))^2+5((2n+1)/(2n-1))^3+...` |
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Answer» Correct Answer - 20 Let `(2n+1)/(2n-1)=r` Then the given series is `S=r+3r^(2)+5r^(3)+7r^(4)+…+(2n-1)r^(n)` `rS=r^(2)+3r^(3)+5r^(4)+..+(2n-3)r^(n)+(2n-1)r^(n+1)` ltbr. Subtracting, we get `(1-r)S=r+2r^(2)+2r^(3)+..+2r^(n)-(2n-1)r^(n+1)` `rArr820(1-r)=r(2r^(2)(1-r^(n-1)))/(1-r)-(2n-1)r^(n+1)` `rArr820(1-r)^(2)=r-r^(2)+2r^(2)-2r^(n+1)-(2n+1)r^(n+1)+(2n-1)r^(n+2)` `rArr820(1-r)^(2)=r+r^(2)-(2n+1)r^(n+1)+(2n-1)r^(n+2)` `rArr820(1-r)^(2)=r+r^(2)-(2n-1)[(2n+1)/(2n-1)r^(n+1)-r^(n+2)]` `rArr820(1-r)^(2)=r+r^(2)-(2n-1)[rcdotr^(n+1)-r^(n+2)]` `rArr820(1-r)^(2)=r(1+r)` `rArr820[1-(2n+1)/(2n-1)]^(2)=(2n+1)/(2n-1)[1+(2n+1)/(2n-1)]` `rArr[(-2)/(2n-1)]^(2)=(2n+1)/(2n-1)[(4n)/(2n-1)]` `rArr820=n(2n+1)` `rArrn=20` |
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