Let `S(x)=1+x-x^(2)-x^(3)+x^(4)+x^(5)-x^(6)-x^(7)+"........+"oo`, where `0ltxlt1`. If `S(x)=(sqrt(2)+1)/(2)`, then the value of `(x+1)^(2)` is

Explanation

`S(x)=1+x-x^(2)-x^(3)+x^(4)+x^(5)-x^(6)-x^(7)+"....."+oo`
where `x in(0,1)`
`S(x)=(1+x)-x^(2)(1+x)+x^(4)+(1+x)-x^(6)(1+x)+"....."+oo`
`impliesS(x)=(1+x)[1-x^(2)+x^(4)-x^(6)+"....."+oo]`
`impliesS(x)=(1+x)((1)/(1+x^(2)))" " [:.S_(oo)=(a)/(1-r) " for "GP]`
According to the question, `S(x)=(sqrt(2)+1)/(2)`
So, `(1+x)/(1+x^(2))=(sqrt(2)+1)/(2)`
`implies=2+2x=(sqrt(2)+1)x^(2)+sqrt(2)+1`
`implies(sqrt(2)+1)x^(2)-2x-2+sqrt(2)+1=0`
`implies(sqrt(2)+1)x^(2)-2x+sqrt(2)-1=0`
`implies(sqrt(2)+1)x^(2)-2x+(1)/(sqrt(2)+1)=0`
`implies[(sqrt(2)+1)x]^(2)-2(sqrt(2)+1)x+1=0`
`implies[(sqrt(2)+1)x-1]^(2)=0`
`implies x=(1)/(sqrt(2)+1)" " ["repeated "]`
So, ` x=sqrt(2)-1`
`:. (x+1)^(2)=2`.