Find the sum of series upto n terms `((2n+1)/(2n-1))+3((2n+1)/(2n-1))^2+5((2n+1)/(2n-1))^3+...`

Answer

20

Explanation

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`