Find the sum of the coefficients of all the integral powers of `x` in the expansion of `(1+2sqrt(x))^(40)dot`

Options

A. `3^(40) +1`
B. `3^(40) -1`
C. `(1)/(2)(3^(40) -1)`
D. `(1)/(2)(3^(40) +1)`

Answer

d

Explanation

We have,
`( 1 + 2sqrt(x))^(40)= ""^(40)C_(0) + ""^(40)C_(1)(2sqrt(x)) + ""^(40)C_(2)(2sqrt(x))^(2) + ...+ ""^(40)C_(40)(2sqrt(x))^(40)`
`rArr ( 1 + 2sqrt(x))^(40)= (""^(40)C_(0) + ""^(40)C_(1)xx2x+ ""^(40)C_(4)xx2^(2) x^(2) + ...+ ""^(40)C_(20)xx 2^(10)x^(10))`
+`(""^(40)C_(1) xx2sqrt(x)+ ""^(40)C_(3)xx2 + ""^(40)C_(3)xx2^(2) + ...+ ""^(40)C_(19)xx 2^(19)x^(10//2))` ...(i)
Putting `sqrt(x) = 1 ` and - 1 respectively, we get
`3^(40) = {""^(40)C_(0)+ ""^(40)C_(2)xx2 + ""^(40)C_(4)xx2^(2) + ...+ ""^(40)C_(20)xx 2^(10)}`
`+{""^(40)C_(1) xx 2 + ""^(40)C_(3)xx2^(3) + ...+ ""^(40)C_(20)xx2^(19)}`
and ,
`1= {""^(40)C_(0) + ""^(40)C_(2)xx2 +""^(40)C_(4)xx2^(2)+...+""^(40)C_(20)xx2^(10)}`
`- {""^(40)C_(1)xx2 + ""^(40)C_(3)xx2^(3) +...+""^(40)C_(19)xx2^(10)}`
`therefore 3^(40) + 1 = 2 {""^(40)C_(0)+ ""^(40)C_(2)xx2+ ""^(40)C_(4) xx2^(2)+...+""^(40)C_(20)xx2^(10)}`
`rArr ""^(40)C_(0)+ ""^(40)C_(2)xx2+ ""^(40)C_(4) xx2^(2)+...+""^(40)C_(20)xx2^(10)= (3^(40) + 1)/(2)` .