Algebra – Binomial Theorem – JEE Main 02 April 2025 Shift 2

Solution:

Given the sum:
$$S = \sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) {}^{11}C_{r+1}$$

Separate the terms in the bracket:
$$S = \sum_{r=0}^{10} \left( \frac{10^{r+1}}{10^r} – \frac{1}{10^r} \right) {}^{11}C_{r+1}$$
$$S = \sum_{r=0}^{10} \left( 10 – \frac{1}{10^r} \right) {}^{11}C_{r+1}$$

Split into two summations:
$$S = 10 \sum_{r=0}^{10} {}^{11}C_{r+1} – \sum_{r=0}^{10} \frac{1}{10^r} {}^{11}C_{r+1}$$

Multiply and divide the second summation by $10$ to match the power with the binomial coefficient index $(r+1)$:
$$S = 10 \sum_{r=0}^{10} {}^{11}C_{r+1} – 10 \sum_{r=0}^{10} {}^{11}C_{r+1} \left(\frac{1}{10}\right)^{r+1}$$

Expand the summations:
$$S = 10 \left[ {}^{11}C_{1} + {}^{11}C_{2} + \dots + {}^{11}C_{11} \right] – 10 \left[ {}^{11}C_{1}\left(\frac{1}{10}\right)^1 + {}^{11}C_{2}\left(\frac{1}{10}\right)^2 + \dots + {}^{11}C_{11}\left(\frac{1}{10}\right)^{11} \right]$$

Using the Binomial expansion $(1+x)^n – 1 = {}^{n}C_{1}x + {}^{n}C_{2}x^2 + \dots + {}^{n}C_{n}x^n$:
$$S = 10 \left[ (1+1)^{11} – 1 \right] – 10 \left[ \left(1 + \frac{1}{10}\right)^{11} – 1 \right]$$

$$S = 10(2^{11} – 1) – 10 \left( \left(\frac{11}{10}\right)^{11} – 1 \right)$$
$$S = 10(2^{11}) – 10 – 10\left(\frac{11^{11}}{10^{11}}\right) + 10$$

$$S = 10 \cdot 2^{11} – \frac{11^{11}}{10^{10}}$$

Multiply and divide the first term by $10^{10}$ to make a common denominator:
$$S = \frac{10 \cdot 2^{11} \cdot 10^{10}}{10^{10}} – \frac{11^{11}}{10^{10}}$$
$$S = \frac{2^{11} \cdot 10^{11}}{10^{10}} – \frac{11^{11}}{10^{10}}$$
$$S = \frac{(20)^{11} – 11^{11}}{10^{10}}$$

Comparing this with the given expression $\frac{\alpha^{11} – 11^{11}}{10^{10}}$:
$$\alpha = 20$$

Ans. (4)