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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Binomial Theorem

If

\sum_{r = 1}^{9} (\frac{r + 3}{2 ^{r}}) \cdot^{9} C_{r} = α (\frac{3}{2})^{9} - β

, where

α, β \in N

, then

(α + β)^{2}

is equal to

(A)27
(B)9
(C)81✓ Correct
(D)18

Answer: (C)

Solution

Split the sum as

\sum_{r = 1}^{9} \frac{r}{2 ^{r}} (r 9) + 3 \sum_{r = 1}^{9} \frac{1}{2 ^{r}} (r 9)

. Using

r (r 9) = 9 (r - 1 8)

, the first part is

\frac{9}{2} \sum_{k = 0}^{8} (k 8) (\frac{1}{2})^{k} = \frac{9}{2} (\frac{3}{2})^{8} = 3 (\frac{3}{2})^{9}

. Since

\sum_{r = 0}^{9} (r 9) (\frac{1}{2})^{r} = (\frac{3}{2})^{9}

, the second part is

3 [(\frac{3}{2})^{9} - 1]

. Combining, the total is

6 (\frac{3}{2})^{9} - 3

, so

α = 6, β = 3

and

(α + β)^{2} = 81

.

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

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