Binomial Expansion?

Question

If ( 1+x )^y = a0 + a1x + a2x^2 + a3x^3 + ... + ayx^y, where y is a positive even number, show that
a0 + a2 + a4 + a6 + ... + ay =( 2^y / 2).

Answer 1

Let x=1 gives 2^y=a0+a1+a2+..+ay
Let x=-1 gives 0=a0-a1+a2+...+ay

Sum gives 2^y=2(a0+a2+..+ay) i. e. result!

Answer 2

The answer can be found out in two steps
Given (1+x)^y = a0+a1.x^1+a2.x^2+a3.x^3 ------------ay.x^y (say equation a)
Step 1
put x=1 in above eqn a) we get
(1+1)^y = a0 +a1(1) + a2(1)^2 -------------ay(1)^y
or 2^y = a0 +a1 +a2+a3+a4------------------ay (equation b)
Step 2
Put x = -1(or minus 1) in equation a we get
(1-1)^y = a0 - a1 + a3 - a4----------- -ay ( even powers of x gives +1 and odd powers gives minus 1 and 0^y = 0)
hence 0 = a0 - a1 + a2 - a3 ------------------------ - ay (equation c)
Taking equation b) and C) together
a0 + a2 +a4+a6--------a(y-1) = a1 +a3 + a5 --------------------ay
= 1/2 (a0 +a1+a2+a3+a4+a5------------ay) = 1/2 (2^y)
Cheers
subhash