# Given the function f(x)=x-4 evaluate the addition of functions f(2), f(2)^2,..f(2)^6 .

### 2 Answers | Add Yours

We have the function f(x) = (x - 4)

f(2), (f(2))^2, (f(2))^3.., form a geometric progression with f(2) = -2 as the first term and f(2) = -2 as the common ratio.

The sum of f(2) + (f(2))^2 + ... (f(2))^6 is given by -2*[(-2)^6 - 1]/(-2 - 1)

=> -2*63/-3

=> 2*21

=> 42

**The required sum is 42.**

We'll have to calculate f(2) + [f(2)]^2 + ... + [f(2)]^6

We'll replace f(2) by it's value => f(2) = 2 - 4 = -2

f(2) + [f(2)]^2 + ... + [f(2)]^6 = -2 + (-2)^2 + (-2)^3 + (-2)^4 + (-2)^5 + (-2)^6

f(2) + [f(2)]^2 + ... + [f(2)]^6 = -2 + 4 - 8 + 16 - 32 + 64

f(2) + [f(2)]^2 + ... + [f(2)]^6 = 2 + 8 + 32

f(2) + [f(2)]^2 + ... + [f(2)]^6 = 42

**The requested sum of the functions is f(2) + [f(2)]^2 + ... + [f(2)]^6 = 42.**