Solve the equation `x^2 -2 = 0`

Explanation

Let `S = (C_(0))/(2) -(C_(1))/(6) + (C_(2))/(10) + (C_(3))/(14) -...+ (-1)^(n) (C_(n))/(4n+2)`
` = 1/2((C_(0))/(1) -(C_(1))/(3) + (C_(2))/(5) - (C_(3))/(7) +...+ (-1)^(n)) ((C_(n))/(2n+1)) `…(i)
Consider , `(1 -x^(2))^(n) = C_(0) - C_(1) x^(2) + C_(2) x^(4) - C_(3)x^(6) + ...+ (-1)^(n) C_(n) x^(2n)`
`rArr int_(0)^(1) (1-x^(2))^(n) dx = int _(0)^(1) (C_(0) - C_(1) x^(2) + C_(2) x^(4) - C_(3) x^(6) + ...+ (-1)^(n) C_(n) x^(2n))dx `
`rArr int_(0)^(1) (1-x^(2))^(n) dx = [C_(0)x - ( C_(1) x^(3))/(3) + (C_(2) x^(5))/(5) -( C_(3) x^(7))/(7) + ...+ (-1)^(n)( C_(n) x^(2n+1))/(2n+1)]`
`rArr int_(0)^(1) (1-x^(2))^(n) dx = [C_(0) - ( C_(1) )/(3) + (C_(2) )/(5) -( C_(3))/(7) + ...+ (-1)^(n)( C_(n) )/(2n+1)]`
From Eq.(i).
`rArr int_(0)^(1) (1-x^(2))^(n) dx= 2S or S = 1/2int_(0)^(1) (1-x^(2))^(n) dx `
Put ` x= sin theta" i.e.,= dx = cos " theta d theta `
`S=1/2 int_(0)^(pi//2) cos^(2n+1) theta d theta `
Bu using Walli's formula,
`S=1/2 (2n(2n-2)(2n-4)...4*2)/((2n+1)(2n-1)(2n-3)...3*1)*1`
`1/2*({2n(2n-2)(2n-4)...4*2}^(2))/((2n+1)!)`
`= 1/2 *((2^(n) n!)^(2))/((2n+1)!) = 2^(2n-1) ((n!)^(2))/((2n+1)!)` .