Strange manipulation with a dummy variable to solve a summation I get sometimes summations where I need to change a number by a dummy variable to integrate/differentiate the sum to get the closed form.
The method works but it seems too strange changing a parameter by a variable to manipulate something and after revert the change to get the solution. An example:
$$\sum_{k\ge 0}k3^k\binom{n}{k}\overset{\text{strange temporal manipulation}}{\longrightarrow}\sum_{k\ge 0}kx^k\binom{n}{k}$$
Then
$$\sum_{k\ge 0}kx^k\binom{n}{k}=x\sum_{k\ge 0}kx^{k-1}\binom{n}{k}=x\frac{\mathrm d}{\mathrm dx}\left(C+\sum_{k\ge 0}x^k\binom{n}{k}\right)=\\=x\frac{\mathrm d}{\mathrm dx}\left(C+(x+1)^n\right)=nx(x+1)^{n-1}$$
Now I change back to get the solution (what is correct)
$$\sum_{k\ge 0}k3^k\binom{n}{k}=3n4^{n-1}$$
Two questions:


*

*We need something special to justify these manipulations?

*There are better methods to solve these sums without the need of this manipulation? Im searching specifically, if exist, some alternative integrate/differentiate that dont need this kind of dummy variable method
Thank you in advance.
 A: I find that something is wrong in your formulas. First, you are asking for the sum with 'k' being negative. I think you meant to be 'k' from zero to 'n'. Then you make a mistake taking out a 'k', which has no sense, because it is the variable that runs inside the summatory. You should have taken an x outside instead.
Finally you obtain $$nk4^{(n-1)}$$ but I think you should have $$3n4^{(n-1)}$$
Answering your questions:
1) The only thing you need to justify is that your substitution can be differentiated. As long as you have $$(x+1)^n$$ it is perfectly differentiable, so you can do this.
2) I think this procedure works pretty well. I don't know any better method. Usually you'll find these kind of summation when dealing with discrete probabilities. You can make a list of common summations to get it done faster, but I think the formal idea works just fine. For sure there should be more ways to get the same result, but let me doubt them to be more straightforward than this. 
You have to be carefull with the variable k. It is different going from zero to n or from 1 to n... Depending the case you are, you will have to extract (or introduce) a term of the summation so you can write the resulting sum.
A: Note that $k{n\choose k}=n{n-1\choose k-1}$. So $$\sum_{k\ge0}k3^k{n\choose k}=\sum_{k\ge0}n3^k{n-1\choose k-1}=3n\sum_{k\ge0}3^{k-1}{n-1\choose k-1}=3n4^{n-1}$$
