With "$n$" being a positive integer, evaluate the sum :
$C(n,0)+2C(n,1)+2^2C(n,2)+...+2^kC(n,k)+...+2^nC(n,n)$
Can someone teach me how to evaluate that using the binomial theorem? The ANSWER is $(1 + 2)^n = 3^n$ but I am not sure how they achieved this answer. Please list out all the steps taken with no shortcuts or left out steps. Step by step explanation if possible please.
Can someone also explain to me how we know that X = 2 and Y = 1 when we use the Binomial theorem to get that answer?
The binomial theorem states that $$(1+x)^n = \sum_{k=0}^n\binom{n}{k}x^k$$ where $n$ is a non negative integer.
If you were to evaluate $x=2$ in the above expression, what would you get?
You seem to have maybe mistyped, by what you said is the answer you must have meant: C(n,0)+2C(n,1)+2^2C(n,2)+...+2^kC(n,k)+...+2^nC(n,n). In anycase this is exactly the binomial theorem,with a=1, b=2, $$(a+b)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\dots + \binom{n}{n}a^{0}b^n.$$ substitue to see that it is the same, $$(1+2)^n=\binom{n}{0}1^n+\binom{n}{1}1^{n-1}2^1+\dots + \binom{n}{n}2^n.$$ The powers of 1 are all just 1. So this tells us that your sum is $(1+2)^n=3^n$.
There are lots of proofs to the binomial theorem if that is what you need, here is one by induction, Proof of $(a+b)^{n+1}$
You can also think about the proof combinatorially by noticing the left hand side of the equation is the number of ways to choose code words where each character can be from a selection of $a$ symbols or from a selection of $b$ different symbols, and symbols can be repeated.
