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Mathematics · Binomial Theorem

JEE Main 2024 — 31 January, Shift 2 — Question 22

Let the coefficient of xrx^{\mathrm{r}} intheexpansionof(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+…….+(x+2)n−1\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}\end{aligned}be αr\alpha_{r}. If ∑r=0nαr=βn−γn,β,γ∈N\sum_{r=0}^{n} \alpha_{r}=\beta^{n}-\gamma^{n}, \beta, \gamma \in N, then the value of β2+γ2\beta^{2}+\gamma^{2} equals \qquad .

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+⋯+(x+2)n−1(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\cdots+(x+2)^{n-1} Let ∑αr\sum \alpha_r be the sum of coefficients. ∑αr=4 n−1+4 n−2⋅3+4 n−3⋅32+⋯+3 n−1\sum \alpha_r=4^{\,n-1}+4^{\,n-2}\cdot 3+4^{\,n-3}\cdot 3^2+\cdots+3^{\,n-1} ∑αr=4 n−1(1+34+(34)2+⋯+(34)n−1)\sum \alpha_r=4^{\,n-1}\left(1+\frac{3}{4}+\left(\frac{3}{4}\right)^2+\cdots+\left(\frac{3}{4}\right)^{n-1}\right) ∑αr=4 n−1(1−(34)n1−34)\sum \alpha_r=4^{\,n-1}\left(\frac{1-\left(\frac{3}{4}\right)^n}{1-\frac{3}{4}}\right) ∑αr=4 n−3 n\sum \alpha_r=4^{\,n}-3^{\,n}

4 n−3 n=βn−γn4^{\,n}-3^{\,n}=\beta^n-\gamma^n β=4,γ=3\beta=4,\qquad \gamma=3 β2+γ2=16+9=25\beta^2+\gamma^2=16+9=25

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Mathematics
Chapter
Binomial Theorem
Topic
Binomial Coefficients