Binomial Theorem Question 13
Question 13 - 2024 (31 Jan Shift 2)
Let the coefficient of $x^{r}$ 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 $\alpha_{r}$. If $\sum_{r=0}^{n} \alpha_{r}=\beta^{n}-\gamma^{n}, \beta, \gamma \in N$, then the value of $\beta^{2}+\gamma^{2}$ equals
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Answer (25)
Solution
$$ \begin{aligned} & (\mathrm{x}+3)^{\mathrm{n}-1}+(\mathrm{x}+3)^{\mathrm{n}-2}(\mathrm{x}+2)+(\mathrm{x}+3)^{\mathrm{n}-3} \ & (\mathrm{x}+2)^{2}+\ldots \ldots+(\mathrm{x}+2)^{\mathrm{n}-1} \ & \sum \alpha_{\mathrm{r}}=4^{\mathrm{n}-1}+4^{\mathrm{n}-2} \times 3+4^{\mathrm{n}-3} \times 3^{2} \ldots \ldots+3^{\mathrm{n}-1} \ & =4^{\mathrm{n}-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^{2} \ldots++\left(\frac{3}{4}\right)^{\mathrm{n}-1}\right] \ & =4^{\mathrm{n}-1} \times \frac{1-\left(\frac{3}{4}\right)^{\mathrm{n}}}{1-\frac{3}{4}} \ & =4^{\mathrm{n}}-3^{\mathrm{n}}=\beta^{\mathrm{n}}-\gamma^{\mathrm{n}}\ & \beta=4, \gamma=3 \ & \beta^{2}+\gamma^{2}=16+9=25 \end{aligned} $$
