If ∑_ r=1 ^9( r+3 2^ r ) . ^9 C_ r =α((3)/(2))^9-β, α, β ∈ N, then (α+β)^2 is equal to

Correct answer: C. 81 ∑_ r=1 ^9( r+3 2^ r ) · ^9 C_ r =α((3)/(2))^9-β, α, β ∈ N Now, aligned & ∑_ r=1 ^9( r+3 2^ r ) · ^9 C_ r =∑_ r=1 ^9( r 2^ r ) · ^9 C_ r +∑_ r=1 ^9( 3 2^ r ) · ^9 C_ r & =∑_ r=1 ^9( 9 2^ r ) · ^8 C_ r-1 +3 ∑_ r=1 ^9 ^9 C_ r ((1)/(2))^ r [U ~g ^9 C_ r ^8 C_ r-1 = 9 r ] & =(9)/(2) ∑_ r=1 ^9 ^8 C_ r-1 ((1)/(2))^ r-1 +3(∑_ r=0 ^9( ^9 C_ r ((1)/(2))^ r )-1) & =(9)/(2) (1+(1)/(2))^8+3((1+(1)/(2))^9-1) aligned =(9)/(2) ·((3)/(2))^8+3((3)/(2))^9-3=6 ·((3)/(2))^9-3 Hence, α=6, β=3 Thus (α+β)^2=81

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Binomial Theorem

If ∑_ r=1 ^9( r+3 2^ r ) . ^9 C_ r =α((3)/(2))^9-β, α, β ∈ N, then...

\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) .{ }^9 \mathrm{C}_{\mathrm{r}}=\alpha\left(\frac{3}{2}\right)^9-\beta, \quad \alpha, \beta \in \mathrm{N}, \quad

then

(\alpha+\beta)^2

is equal to

Options

A $27$
B $9$
C $81$
D $18$

Correct answer

C. $81$

Step-by-step solution

\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}=\alpha\left(\frac{3}{2}\right)^9-\beta, \alpha, \beta \in \mathrm{N}

Now,

\begin{aligned} & \sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}+3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}=\sum_{\mathrm{r}=1}^9\left(\frac{\mathrm{r}}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}}+\sum_{\mathrm{r}=1}^9\left(\frac{3}{2^{\mathrm{r}}}\right) \cdot{ }^9 \mathrm{C}_{\mathrm{r}} \\ & =\sum_{\mathrm{r}=1}^9\left(\frac{9}{2^{\mathrm{r}}}\right) \cdot{ }^8 \mathrm{C}_{\mathrm{r}-1}+3 \sum_{\mathrm{r}=1}^9{ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\left[\mathrm{U} \sin \mathrm{~g} \frac{{ }^9 \mathrm{C}_{\mathrm{r}}}{{ }^8 \mathrm{C}_{\mathrm{r}-1}}=\frac{9}{\mathrm{r}}\right] \\ & =\frac{9}{2} \sum_{\mathrm{r}=1}^9{ }^8 \mathrm{C}_{\mathrm{r}-1}\left(\frac{1}{2}\right)^{\mathrm{r}-1}+3\left(\sum_{\mathrm{r}=0}^9\left({ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}\right)-1\right) \\ & =\frac{9}{2}\left(1+\frac{1}{2}\right)^8+3\left(\left(1+\frac{1}{2}\right)^9-1\right) \end{aligned}

=\frac{9}{2} \cdot\left(\frac{3}{2}\right)^8+3\left(\frac{3}{2}\right)^9-3=6 \cdot\left(\frac{3}{2}\right)^9-3

Hence,

\alpha=6, \beta=3

Thus

(\alpha+\beta)^2=81

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