Let X = ( 10 C 1 ) 2 +2 ( 10 C 2 ) 2 +3 ( 10 C 3 ) 2 +ldots . .+10 ( 10 C 10 ) 2 where 10 C r , r in\ 1,2, ldots . ., 10\ denote binomial coefficients. Then the value of 1/1430 X is \ \ \ \ .

Mathematics · Binomial Theorem

JEE Advanced 2018 — Paper 2 — Question 41

Let $\mathrm{X}=\left({ }^{10} \mathrm{C}_{1}\right)^{2}+2\left({ }^{10} \mathrm{C}_{2}\right)^{2}+3\left({ }^{10} \mathrm{C}_{3}\right)^{2}+\ldots . .+10\left({ }^{10} \mathrm{C}_{10}\right)^{2}$ where ${ }^{10} \mathrm{C}_{\mathrm{r}}, \mathrm{r} \in\{1,2, \ldots . ., 10\}$ denote binomial coefficients. Then the value of $\frac{1}{1430} \mathrm{X}$ is $\_\_\_\_$ .

Answer: 646

Numerical answer — enter this value.

Step-by-step solution

The sum is $X = \sum_{r=1}^{10} r \cdot \binom{10}{r}^2$ . The standard identity: $\sum_{r=1}^{n} r \cdot \binom{n}{r}^2 = n \cdot \binom{2n-1}{n-1}$ .

For $n=10$ , $X = 10 \cdot \binom{19}{9}$ . Compute $\binom{19}{9} = \frac{19!}{9!10!} = 92378$ . Thus $X = 10 \times 92378 = 923780$ . The problem asks for $\frac{X}{1430}$ . Simplify: $\frac{923780}{1430} = 646$ . Therefore, the required value is $646$ .

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Series involving sum & product of Binomial Coefficients