Simplify \sum_{i=0}^{n}\binom{n}{i}2^{n-i} to 3^n

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The discussion focuses on simplifying the expression \sum_{i=0}^{n}\binom{n}{i}2^{n-i} to demonstrate that it equals 3^n. The original equation was rewritten, but the poster struggled with the next steps due to the changing combinatorial and powered terms. A hint was provided, referencing the binomial expansion of (2 + x)^n. The poster realized the solution after considering this hint. The problem was successfully resolved with this insight.
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Homework Statement



This is actually part of a larger problem that asks us to prove that the number of ways of counting something is equal to 3^n. I have worked it out and the equation I get is:

\binom{n}{0}2^n + \binom{n}{1}2^{n-1}+\ldots+\binom{n}{n}2^{n-n}

I am wondering how I should simplify this to make it equal to 3^n

2. The attempt at a solution

I rewrote the above equation into:

\displaystyle\sum_{i=0}^{n}\binom{n}{i}2^{n-i}

But then I didn't know how to proceed from here since both the combinatorial choosing term and the powered terms are changing. I also tried factoring out 2^n but that didn't do anything.

Can anyone help me?
Thanks.
 
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Hi noblerare! :smile:

Hint: what's (2 + x)n ? :wink:
 
Ohhhh, wow. Okay thanks, tiny-tim! Problem solved.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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