JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 28)
Let the coefficient of $$x^r$$ in the expansion of $$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots \ldots \ldots .+(x+2)^{n-1}$$ be $$\alpha_r$$. If $$\sum_\limits{r=0}^n \alpha_r=\beta^n-\gamma^n, \beta, \gamma \in \mathbb{N}$$, then the value of $$\beta^2+\gamma^2$$ equals _________.
Answer
25
Explanation
$$\begin{aligned}
& (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} \\
& (x+2)^2+\ldots \ldots .+(x+2)^{n-1} \\
& \sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} \\
& =4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots .+\left(\frac{3}{4}\right)^{n-1}\right] \\
& =4^{n-1} \times \frac{1-\left(\frac{3}{4}\right)^n}{1-\frac{3}{4}} \\
& =4^n-3^n=\beta^n-\gamma^n \\
& \beta=4, \gamma=3 \\
& \beta^2+\gamma^2=16+9=25
\end{aligned}$$
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