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Let the coefficient of $x^{\mathrm{r}}$ in the expansion of $(\mathrm{x}+3)^{\mathrm{n}-1}+(\mathrm{x}+3)^{\mathrm{n}-2}(\mathrm{x}+2)+$ $(\mathrm{x}+3)^{\mathrm{n}-3}(\mathrm{x}+2)^2+\ldots \ldots+(\mathrm{x}+2)^{\mathrm{n}-1}$ be $\alpha_{\mathrm{r}}$. If $\sum_{\mathrm{r}=0}^{\mathrm{n}} \alpha_{\mathrm{r}}=\beta^{\mathrm{n}}-\gamma^{\mathrm{n}}, \beta, \gamma \in \mathrm{N}$, then the value of $\beta^2+\gamma^2$ equals..................
$23$
$24$
$20$
$25$
Solution
$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} $
$ (x+2)^2+\ldots \ldots . .+(x+2)^{n-1} $
$\sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} $
$=4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots . .+\left(\frac{3}{4}\right)^{n-1}\right]$
$=4^{\mathrm{n}-1} \times \frac{1-\left(\frac{3}{4}\right)^{\mathrm{n}}}{1-\frac{3}{4}} $
$=4^{\mathrm{n}}-3^{\mathrm{n}}=\beta^{\mathrm{n}}-\gamma^{\mathrm{n}} $
$ \beta=4, \gamma=3 $
$\beta^2+\gamma^2=16+9=25$
