JEE Main 2022 (Online) 25th June Morning ShiftQuestion 1
Let $C_r$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+x)^{10}$. If for $\alpha,\beta\in\mathbf{R}$, $C_1+3\cdot2C_2+5\cdot3C_3+\ldots$ upto $10$ terms $=\frac{\alpha\times2^{11}}{2^{\beta}-1}\bigg(C_0+\frac{C_1}{2}+$ $\frac{C_2}{3}+\ldots\text{ upto }10\text{ terms}\bigg)$ then the value of $\alpha+\beta$ is equal to $\_\_\_\_\_\_$ .
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$286$
Solution
Concept. The left side is $\sum(2k-1)k\,C_k$; the bracket on the right is $\sum\frac{C_k}{k+1}$. Formulas used. (1) $\sum k\,C_k=n\,2^{n-1}$ and $\sum k^2\,C_k=n(n+1)\,2^{n-2}$(2) $\sum_{k=0}^{n}\dfrac{C_k}{k+1}=\dfrac{2^{n+1}-1}{n+1}$Step 1 - Left side. The $k$-th term is $(2k-1)\,k\,C_k$ for $k=1,\dots,10$. By (1) with $n=10$: $$2\cdot10\cdot11\cdot2^8-10\cdot2^9=56320-5120=51200.$$ Step 2 - The bracket. Taking the bracket as $C_0+\frac{C_1}{2}+\cdots+\frac{C_{10}}{11}$, by (2) it equals $\dfrac{2^{11}-1}{11}=\dfrac{2047}{11}$. Step 3 - Solve for $\frac{\alpha}{2^\beta-1}$. $$51200=\frac{\alpha\cdot2^{11}}{2^\beta-1}\cdot\frac{2047}{11}\ \Longrightarrow\ \frac{\alpha}{2^\beta-1}=\frac{51200\cdot11}{2048\cdot2047}=\frac{275}{2047}.$$ Since $2047=2^{11}-1$: $\alpha=275$, $\beta=11$. Step 4 - Value. $\alpha+\beta=286$. Answer: 286