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

Question

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.

Answer 1

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 $$\binom{n+1}{2}(3^{2}-2^{2})=5\binom{n+1}{2}$$

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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 $3n^{2}$.