JEE Main 2025 (Online) 2nd April Evening ShiftQuestion 1
If $\sum_{r=0}^{10}\left(\frac{10^{r+1}-1}{10^{r}}\right)\cdot{}^{11}C_{r+1}=\frac{\alpha^{11}-11^{11}}{10^{10}}$, then $\alpha$ is equal to :
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. $\dfrac{10^{r+1}-1}{10^r}=10-10^{-r}$; each part is a binomial sum. Formulas used. (1) $\sum_{j=0}^{n}{}^{n}C_j\,x^j=(1+x)^n$Step 1 - Re-index. Put $j=r+1$ ($j=1,\dots,11$); then $10-10^{-r}=10-10^{1-j}$ and $$\text{sum}=10\sum_{j=1}^{11}{}^{11}C_j-10\sum_{j=1}^{11}{}^{11}C_j\,10^{-j}.$$ Step 2 - Use (1). The $j=0$ term is missing from both sums: $$\text{sum}=10\left(2^{11}-1\right)-10\left[\left(\frac{11}{10}\right)^{11}-1\right]=10\cdot2^{11}-\frac{11^{11}}{10^{10}}.$$ Step 3 - Common denominator. $10\cdot2^{11}=\dfrac{2^{11}\cdot10^{11}}{10^{10}}=\dfrac{20^{11}}{10^{10}}$, so $$\text{sum}=\frac{20^{11}-11^{11}}{10^{10}}.$$ So $\alpha=20$. Answer: (C)