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7.Binomial Theorem
hard
Let $X =\left({ }^{10} C _1\right)^2+2\left({ }^{10} C _2\right)^2+3\left({ }^{10} C _3\right)^2+\ldots \ldots . .+10\left({ }^{10} C _{10}\right)^2$ where ${ }^{10} C _{ r }, r \in\{1,2, \ldots ., 10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is. . . . . . .
A
$430$
B
$435$
C
$540$
D
$646$
(IIT-2018)
Solution
$X=\sum_{r=0}^{10} r\left({ }^{10} C_r\right)^2$
$X=\sum_{r=0}^{10}(10-r)\left({ }^{10} C_{10-r}\right)^2$
$2 X=\sum_{r=0}^{10} 10\left({ }^{10} C_r\right)^2$
$X=5 \cdot{ }^{20} C_{10} \Rightarrow \frac{X}{1430}=646.00$
Standard 11
Mathematics
