Q. 1) Binomial Theorem.

Fill in the blank. (Binomial Theorem)
If (1 + x)^n = C(0) + C(1) * x + C(2) * x^2 + . . . . . . + C(n) * x^n, then

(1/2) * C(1) + (1/4) * C(3) + (1/6) * C(5) + . . . is equal to - - - - - - - - - -

[Where, C(1) , C(2) , C(3) . . . C(n) are binomial coefficients ]

SOLUTION ► (2^n - 1) / (n + 1) Answer.

Integrating it from 0 to 1, and from (-1) to 0 and
subtracting second from first result we get,

∫(0 to 1) (1 + x)^n dx = ∫(0 to 1) [ C(0) + C(1) * x + C(2) * x^2 + . . + C(n) * x^n ] dx
=> [2^(n + 1) - 1]/(n + 1) = C(0)/1 + C(1)/2 + C(2)/3 + C(3)/4 + C(4)/5 .+ .+ . - - - - (i)

∫(-1 to 0) (1 + x)^n dx = ∫(-1 to 0) [ C(0) + C(1) * x + C(2) * x^2 + . . + C(n) * x^n ] dx
=>1/(n + 1) = C(0)/1 - C(1)/2 + C(2)/3 - C(3)/4 + C(4)/5 - C(5)/6.+ .+ . - - - - (ii)
=> use, (ii) - (i) and divide by 2,

=> C(1)/2 + C(3)/4 + C(5)/6 + C(7)/8 + .. + ..+ = [2^n - 1]/(n + 1) ► (Ans)