7.Binomial Theorem
hard

माना $(\mathrm{x}+3)^{\mathrm{n}-1}+(\mathrm{x}+3)^{\mathrm{n}-2}(\mathrm{x}+2)+$ $(x+3)^{n-3} \cdot(x+2)^2+\ldots \ldots .+(x+2)^{n-1}$ के प्रसार में $x^r$ का गुणांक $\alpha_r$ है। यदि $\sum_{\mathrm{r}=0}^{\mathrm{n}} \alpha_{\mathrm{r}}=\beta^{\mathrm{n}}-\gamma^{\mathrm{n}}, \beta, \gamma \in \mathrm{N}$ है, तो $\beta^2+\gamma^2$ बराबर है ........

A

$23$

B

$24$

C

$20$

D

$25$

(JEE MAIN-2024)

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 $

$\beta^2+\gamma^2=16+9=25$

Standard 11
Mathematics

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