Find $k$ if $f(x)=k(1/2)^x$ for $x=1,2,3,$ and $0$ elsewhere.

Question

Find $k$ if $f(x)=k(1/2)^x$ for $x=1,2,3,$ and $0$ elsewhere.

Here is what I did:

$$\int_1^3 k\left(\frac12\right)^{x}dx=1$$ After integrating, I found it to be:

$$\left[\frac{-k}{ln(2)2^x}\right]_1^{3}=1$$ When I evaluated this I got $k=1.848$...which appears to be wrong as I was told $k=\frac87$

Did I do the integral wrong? How would you find the integral of $(\frac12)^x$? Or, is there some other error there?

Answer 1

Hint: $$f(1)+f(2)+f(3)=k\,\left(\frac12+\frac14+\frac18\right)=k\,\frac78$$