Verified Solution • 2025–26 Session moderate QJEEM24Ma3d5c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma3d5c Solution: Let the coefficient of in the expansion of be . If ∑ r = 0 n α… Let the coefficient of xrx^rxr in the expansion of (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+\ldots \ldots \ldots .+(x+2)^{n-1}(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+……….+(x+2)n−1 be αr\alpha_rαr . If ∑r=0nαr=βn−γn,β,γ∈N, then the value of β2+γ2\beta^2+\gamma^2β2+γ2 equals _________. Step-by-Step Verified Solution amp;(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3amp;(x+2)2+…….+(x+2)n−1amp;∑αr=4n−1+4n−2×3+4n−3×32……+3n−1amp;=4n−1[1+34+(34)2….+(34)n−1]amp;=4n−1×1−(34)n1−34amp;=4n−3n=βn−γnamp;β=4,γ=3amp;β2+γ2=16+9=25\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}amp;(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3amp;(x+2)2+…….+(x+2)n−1amp;∑αr=4n−1+4n−2×3+4n−3×32……+3n−1amp;=4n−1[1+43+(43)2….+(43)n−1]amp;=4n−1×1−431−(43)namp;=4n−3n=βn−γnamp;β=4,γ=3amp;β2+γ2=16+9=25 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring