Question

$f(n)=\sum_{k=0}^{n} \binom{n+k}{k} \frac{1}{2^k}$; then find $\frac{f(12)+f(14)}{f(9)}= \dots$

Answer 1

40

Answer 2

We are given

$$f(n)=\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}.$$

It is a known result (verified by testing small values, e.g., $n=0,1,2,3$) that

$$f(n)=2^n.$$

For example, when $n=1$:

$$f(1)=\binom{1}{0}\left(\frac{1}{2}\right)^0+\binom{2}{1}\left(\frac{1}{2}\right)^1=1+2\cdot\frac{1}{2}=1+1=2=2^1.$$

Thus,

$$f(12)=2^{12},\quad f(14)=2^{14},\quad f(9)=2^9.$$

Now, compute the required expression:

$$\frac{f(12)+f(14)}{f(9)}=\frac{2^{12}+2^{14}}{2^9}=\frac{2^{12}}{2^9}+\frac{2^{14}}{2^9}=2^{3}+2^{5}=8+32=40.$$

Core Explanation:
Recognize the identity $f(n)=\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^n$. Substitute $n=12, 14, 9$ and simplify.