JEE Advanced 2018 Paper 2Question 1
Let $$X=\left({}^{10}C_1\right)^2+2\left({}^{10}C_2\right)^2+3\left({}^{10}C_3\right)^2+\cdots+10\left({}^{10}C_{10}\right)^2,$$ where ${}^{10}C_r,\ r\in\{1,2,\cdots,10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430}X$ is _____ .
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$646$
Solution
Concept. Absorb $r$ into one coefficient; the sum becomes a Vandermonde convolution. Formulas used. (1) $r\,{}^{n}C_r=n\,{}^{n-1}C_{r-1}$(2) Vandermonde: $\sum_r{}^{9}C_{r-1}\,{}^{10}C_{10-r}={}^{19}C_9$Step 1 - Absorb $r$. By (1), and writing the second ${}^{10}C_r$ as ${}^{10}C_{10-r}$: $$X=10\sum_{r=1}^{10}{}^{9}C_{r-1}\,{}^{10}C_{10-r}.$$ Step 2 - Vandermonde. The lower indices add up to $9$, so by (2), $X=10\,{}^{19}C_9=10\cdot92378=923780$. Step 3 - Value. $\dfrac{923780}{1430}=646$. Answer: 646