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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Binomial Theorem

If

\sum_{r = 0}^{10} (\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}

, then

α

is equal to :

(A) $15$
(B) $11$
(C) $24$
(D) $20$ ✓ Correct

Answer: (D)

Solution

Since

\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}} = 10 - 1 0^{- r}

, the sum is

10 \sum_{r = 0}^{10}^{11} C_{r + 1} - \sum_{r = 0}^{10} 1 0^{- r}^{11} C_{r + 1}

. The first part is

10 \sum_{j = 1}^{11}^{11} C_{j} = 10 (2^{11} - 1)

. For the second, with

j = r + 1

,

\sum_{r = 0}^{10} 1 0^{- r}^{11} C_{r + 1} = 10 \sum_{j = 1}^{11}^{11} C_{j} 1 0^{- j} = 10 [(1 + \frac{1}{10})^{11} - 1] = \frac{1 1 ^{11}}{1 0 ^{10}} - 10

. Combining, the sum is

10 \cdot 2^{11} - \frac{1 1 ^{11}}{1 0 ^{10}}

. Since

10 \cdot 2^{11} = \frac{2 ^{11} \cdot 1 0 ^{11}}{1 0 ^{10}} = \frac{2 0 ^{11}}{1 0 ^{10}}

, the sum equals

\frac{2 0 ^{11} - 1 1 ^{11}}{1 0 ^{10}}

, so

α = 20

.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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