Question

the sum of the coefficients of even powers of x of a polynomial f (x) is

Answer 1

Answer:

0.5(f(1)+f(-1))

Step-by-step explanation:

f(x)=a0+a1*x^1+....+an*x^n

n is odd

f(1)=a0+a1+.......an

f(-1)=a0-a1+a2+.................+a(n-2)-a(n-1)

f(1)+f(-1)=2(sum of coefficients of even powers of x)

sum of even coefficients of x=0.5(f(1)+f(-1))

Answer 2

To find: The sum of the coefficients of even powers of x of a polynomial f (x)

Step-by-step explanation:

let the polynomial be f(x)

To find the sum of the coefficients of even powers of x of a polynomial f (x) is

the expansion of the polynomial

$f(x)= a_0 +a_1 x^1 +a_2 x_2.....a_n x^n\\\\\text{put n =1 and n = -1 then }\\\\f(1)= a_0+a_1+.....a_n..(1)$

$\\\\f(-1)=a_0-a_1+a_2+....a(n-2)-a(n-1)...(2)\\\\\text{add 1 and 2 we get}\\\\f(1)+f(-1)= 2(\text{sum of coefficient of even power of x})\\\\then \\\\\text{sum of coefficient of even power of x} = \frac{1}{2}(f(1)+f(-1))$

#Learn more :

https://brainly.in/question/14336859