Using the binomial expansion to show $\sum_{k=0}^n {n \choose k} \frac{1}{2^k} = (3/2)^n$ I have to show $$\sum_{k=0}^n  {n \choose k} \frac{1}{2^k} = (3/2)^n$$ using the binomial theorem. I haven't had any practice on these types of questions so i'm unsure as to how to proceed. Would anybody be able to what the method is for solving these types of questions, and, is there any general method of what we're trying to do? What's the thought process to solve these types of questions
Much appreciated.
Thank you
 A: $\left(\dfrac{3}{2}\right)^n= \left(\dfrac{1}{2}+1\right)^n =\sum\limits_{k=0}^n\dbinom{n}{k}\dfrac{1}{2^k}$
A: Take $x=1/2$ and $y=1$ in the expansion 
$$
(x+y)^n= \sum_{k=0}^n \binom{n}{k}x^k y^{n-k}.
$$
A: $$(3/2)^n=(1+1/2)^n=\sum_{k=0}^n  {n \choose k} \frac{1}{2^k}.$$
Last equation follows from binomial theorem. 
A: There is a theorem (Binomial theorem, demostrable by induction) that shows you:
\begin{equation}
(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}
\end{equation}
Then if you use the $\displaystyle a=\frac{1}{2}$ and $b=1$, you obtain:
$$\sum_{k=0}^n  \binom{n}{k} \frac{1}{2^k} = (3/2)^n$$
Proof:
The demostration of the Binomial Theorem is not complicated, the case $n=1$ is obviously true, then if we assume the case $n$, and we came to prove the case $n+1$ and the theorem is proven. If:
$$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$$
Then:
\begin{eqnarray*}
(a+b)^{n+1} &=& (a+b)(a+b)^{n} = a(a+b)^{n} + b(a+b)^{n} \\
(a+b)^{n+1} &=& a\sum_{k=0}^n \binom{n}{k} a^k b^{n-k} + b\sum_{k=0}^n \binom{n}{k} a^k b^{n-k}  \\
(a+b)^{n+1} &=& \sum_{k=0}^n \binom{n}{k} a^{k+1} b^{n-k} + \sum_{k=0}^n \binom{n}{k} a^k b^{n-k+1}  \\
(a+b)^{n+1} &=& \sum_{k=1}^{n+1} \binom{n}{k-1} a^{k} b^{n-k+1} + \sum_{k=0}^n \binom{n}{k} a^k b^{n-k+1}  \\
(a+b)^{n+1} &=& \left[a^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{k} b^{n-k+1} \right] + \left[b^{n+1} +\sum_{k=1}^n \binom{n}{k} a^k b^{n-k+1} \right] \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \binom{n}{k-1} + \binom{n}{k}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \frac{n!}{(k-1)!(n-k+1)!} + \frac{n!}{k!(n-k)!}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \frac{n!k}{k!(n-k+1)!} + \frac{n!(n-k+1)}{k!(n-k+1)!}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left(a^k b^{n-k+1}\left[ \frac{n!(n+1)}{k!(n-k+1)!}\right]\right) \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\left[a^k b^{n-k+1} \frac{(n+1)!}{k!(n+1-k)!}\right] \\
(a+b)^{n+1} &=& a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\binom{n+1}{k}a^k b^{n+1-k} \\
(a+b)^{n+1} &=& \sum_{k=0}^{n+1}\binom{n+1}{k}a^k b^{n+1-k} \\
\end{eqnarray*}
And now the theorem is demostrated by induction. Here is a video of the demonstration.
