the coefficient of $ x^{n-1}$ in $(x+3)^n+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^2\cdots(x+2)^n$ 
Find the coefficient of $ x^{n-1}$ in:
   $\ \\ (x+3)^n+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^2\cdots+(x+2)^n$   

My Approach:
 $(x+3)^n+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^2\cdots+(x+2)^n=\sum(x+3)^{n-r}(x+2)^r$
For each term of this summation:
we have to calculate the number of ways of choosing one (x+a) for which we would multiply the 'a' and not the 'x' part. They can be chosen in $\binom{n-r}{1}+\binom{r}{1}$ ways. In case the chosen is (x-3) the coefficient of $x^{n-1}$ would be 3 else it would be 2. Hence the coefficient of $x^{n-1}$ for the general term would be $3\binom{n-r}{1}+2\binom{r}{1}$.   
This simplifies to $3(n-r)+2r=3n-r$. The coefficient of $x^{n-1}$ in the summation is $\sum_0^n 3n-\sum_{r=0}^n{n} r=3n^2-\frac{n(n+1)}{2}=\frac{6n^2-n^2-n}{2}=\frac{5n^2-n}{2}$.    
The answer given is $5\binom{n+1}{2}$ but I can't figure out what the error is in this logic. It would be great if someone could help me answer this question.

It is easier to simplify the given expression by using formula $$a^{r-1}+a^{r-2}b+\cdots+ab^{r-2} +b^{r-1}=\frac{a^{r}-b^{r}}{a-b}$$ which is easily proved by multiplying both sides of the equation by $a-b$. Here we have $r=n+1,a=x+3,b=x+2$ and hence the expression simplifies to $(x+3)^{n+1}-(x+2)^{n+1}$ and the desired coefficient is $$\binom{n+1}{2}(3^{2}-2^{2})=5\binom{n+1}{2}$$

You had your approach correct but you made a calculation mistake while calculating the sum $\sum_{r=0}^{n}3n$ which should evaluate to $3n(n+1)$ but you wrote $3n^{2}$.
