Binomial Theorem
Geometric Series / Coefficient via GP Summation
nta_pyq_2024_jan
Grade 11
Question:
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 $\displaystyle\sum_{r=0}^{n}\alpha_r = \beta^n - \gamma^n$, $\beta,\gamma\in\mathbb{N}$, then the value of $\beta^2+\gamma^2$ equals ______.
Step-by-Step Solution
Key Concept: Recognize the sum as a geometric series with ratio $\frac{x+2}{x+3}$. Using the GP sum formula: $\sum = \frac{(x+3)^n-(x+2)^n}{(x+3)-(x+2)}=(x+3)^n-(x+2)^n$. Then $\sum_{r=0}^n \alpha_r$ is obtained by putting $x=1$.
Sum $= (x+3)^{n-1}\cdot\frac{1-(\frac{x+2}{x+3})^n}{1-\frac{x+2}{x+3}} = (x+3)^n-(x+2)^n$.
$\sum_{r=0}^n\alpha_r=$ put $x=1$: $4^n-3^n=\beta^n-\gamma^n$.
So $\beta=4,\gamma=3$ and $\beta^2+\gamma^2=16+9=25$.
Correct Answer: 25