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!
Best Answer
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*}