Find the coefficients of x^r where 0 le r le | vedclass

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(B) The given expression is a geometric series with first term $a = (x + 3)^{n - 1}$,common ratio $q = \frac{x + 2}{x + 3}$,and $n$ terms.
Using the

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$^nC_r(3^{n - r} - 2^{n - r})$

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(B) The given expression is a geometric series with first term $a = (x + 3)^{n - 1}$,common ratio $q = \frac{x + 2}{x + 3}$,and $n$ terms.
Using the sum formula for a geometric series $S_n = \frac{a(1 - q^n)}{1 - q}$:
$S_n = \frac{(x + 3)^{n - 1} [1 - (\frac{x + 2}{x + 3})^n]}{1 - \frac{x + 2}{x + 3}} = \frac{(x + 3)^{n - 1} [\frac{(x + 3)^n - (x + 2)^n}{(x + 3)^n}]}{\frac{(x + 3) - (x + 2)}{x + 3}} = \frac{(x + 3)^n - (x + 2)^n}{(x + 3) - (x + 2)} = (x + 3)^n - (x + 2)^n$.
Now,we find the coefficient of $x^r$ in the expansion of $(x + 3)^n - (x + 2)^n$.
Using the Binomial Theorem,$(x + a)^n = \sum_{k=0}^{n} {^nC_k} x^{n-k} a^k$.
For $(x + 3)^n$,the term containing $x^r$ is $^nC_{n-r} x^r 3^{n-r} = ^nC_r x^r 3^{n-r}$.
For $(x + 2)^n$,the term containing $x^r$ is $^nC_{n-r} x^r 2^{n-r} = ^nC_r x^r 2^{n-r}$.
Thus,the coefficient of $x^r$ is $^nC_r 3^{n-r} - ^nC_r 2^{n-r} = ^nC_r(3^{n - r} - 2^{n - r})$.

Find the coefficients of $x^r$ where $0 \le r \le (n - 1)$ in the expansion of $(x + 3)^{n - 1} + (x + 3)^{n - 2}(x + 2) + (x + 3)^{n - 3}(x + 2)^2 + ... + (x + 2)^{n - 1}$.

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