Let the coefficient of x^{r} in the expansion | vedclass

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

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$25$

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(D) The given expression is a geometric series with first term $a = (x+3)^{n-1}$,common ratio $r = \frac{x+2}{x+3}$,and $n$ terms.Using the sum formula $S_n = a \frac{1-r^n}{1-r}$,we get:$S = (x+3)^{n-1} \frac{1 - (\frac{x+2}{x+3})^n}{1 - \frac{x+2}{x+3}} = (x+3)^{n-1} \frac{1 - \frac{(x+2)^n}{(x+3)^n}}{\frac{x+3-x-2}{x+3}} = (x+3)^{n-1} \frac{(x+3)^n - (x+2)^n}{(x+3)^n} 	imes (x+3) = (x+3)^n - (x+2)^n$.The sum of the coefficients $\sum \alpha_r$ is obtained by setting $x=1$ in the expression $(x+3)^n - (x+2)^n$.$\sum \alpha_r = (1+3)^n - (1+2)^n = 4^n - 3^n$.Given $\sum \alpha_r = \beta^n - \gamma^n$,we have $\beta = 4$ and $\gamma = 3$.Therefore,$\beta^2 + \gamma^2 = 4^2 + 3^2 = 16 + 9 = 25$.

Let the coefficient of $x^{r}$ in the expansion of $(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots+(x+2)^{n-1}$ be $\alpha_{r}$. If $\sum_{r=0}^{n-1} \alpha_{r}=\beta^{n}-\gamma^{n}$,where $\beta, \gamma \in N$,then the value of $\beta^2+\gamma^2$ equals:

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