Binomial Theorem Question 13
Question 13 - 2024 (31 Jan Shift 2)
Let the coefficient of $x^{r}$ in the expansion of $(x+3)^{n-1}+(x+3)^{n-2}(x+2)+$ $(x+3)^{n-3}(x+2)^{2}+\ldots \ldots+(x+2)^{n-1}$ 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
$$ \begin{aligned} & (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} \\ & (x+2)^{2}+\ldots \ldots+(x+2)^{n-1} \\ & \sum \alpha _r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^{2} \ldots \ldots+3^{n-1} \\ & =4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^{2} \ldots++\left(\frac{3}{4}\right)^{n-1}\right] \\ & =4^{n-1} \times \frac{1-\left(\frac{3}{4}\right)^{n}}{1-\frac{3}{4}} \\ & =4^{n}-3^{n}=\beta^{n}-\gamma^{n}\ & \beta=4, \gamma=3 \\ & \beta^{2}+\gamma^{2}=16+9=25 \end{aligned} $$
