Alg10.18

Find the sum of the coefficients of the even powers of x in the expansion of $(1+x+x^2+x^3)^5\,$

Let $(1+x+x^2+x^3)^5=a_0+a_1 x+a_2 x^2+......+a_{15} x^{15}\,$

where $a_0,a_1.....a_{15}\,$ are coefficients. Substituting $x=1\,$

$4^5=a_0+a_1+a_2+.............+a_{15}\,$ .Let this equation be 1.

Substituting $x=-1\,$

$0=a_0-a_1+a_2-........-a_{15}\,$ Let this be equation 2.

Adding 1 and 2,we get

$4^5=2(a_0+a_2+a_4+......+a_{14})\,$

Therefore $a_0+a_2+a_4+.......+a_{14}=\frac{4^5}{2}=\frac{1024}{2}=512\,$

The sum of the coefficients of the ven powers of $x\,$ is $512\,$