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By setting $x$ equal to the appropriate values in the binomial expansion (or one of its derivatives, etc.), evaluate

$$ \sum_{k=1}^n k3^k {n \choose k} $$

Basically this problem is tasking me with taking the Binomial Theorem:

$$ (1+x)^n = \sum_{k=0}^n {n \choose k} x^k $$

And performing manipulations until it reaches the given form, in which case a value of $x$ can be determined.

By taking the derivative, we have that:

$$ n(1+x)^{n-1} = \sum_{k=0}^n {n \choose k} kx^{k-1} $$

By changing the bounds:

$$ n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} (k-1)x^{(k-1)-1} $$

$$ n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} (k-1)\frac{x^k}{x^2} $$

This could also become:

$$ x^2n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} (k-1)x^k $$

Which is pretty close to the desired form, but I'm not really sure what to do with the $k-1$ term inside ${n \choose k-1}$. I can further simplify by doing:

$$ x^2n(1+x)^{n-1} = \sum_{k=1}^n {n \choose k-1} kx^k - x^k $$

And I'm not sure if this is a valid simplification, given the Sigma:

$$ x^2n(1+x)^{n-1} + x^k = \sum_{k=1}^n {n \choose k-1} kx^k $$

How can I achieve this desired form? Thanks so much! I appreciate the help!

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    $\begingroup$ I think multiply by $x$ the both sides of $$ n(1+x)^{n-1} = \sum_{k=0}^n {n \choose k} kx^{k-1} $$ would work. $\endgroup$ – Cave Johnson Sep 20 '16 at 5:34
  • $\begingroup$ A combinatorial way to get this through is by using $k\binom nk = n \binom{n-1}{k-1}$. $\endgroup$ – Sungjin Kim Sep 20 '16 at 5:58
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Hint: Note that changing the bounds give \begin{align*} \sum_{k=0}^n\binom{n}{k}kx^{k-1}=\sum_{k=1}^{n\color{blue}{+1}}\binom{n}{k-1}(k-1)x^{(k-1)-1} \end{align*}

Another convenient way to calculate the sum is using the binomial identity \begin{align*} k\binom{n}{k}=n\binom{n-1}{k-1} \end{align*}

We obtain \begin{align*} \sum_{k=1}^nkx^k\binom{n}{k}&=n\sum_{k=1}^n\binom{n-1}{k-1}x^{k}\\ &=nx\sum_{k=0}^{n-1}\binom{n-1}{k}x^{k}\\ &=nx(1+x)^{n-1}\\ \end{align*}

Evaluating the expression at $x=3$ gives the result.

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Consider $$\sum_{k=1}^n k {n \choose k}x^k=x\sum_{k=1}^n k {n \choose k}x^{k-1}=x \left(\sum_{k=1}^n {n \choose k}x^{k} \right)'=x\left((1+x)^n-1\right))'=nx(1+x)^{n-1}$$

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