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7.Binomial Theorem
hard
माना कि $X=\left({ }^{10} C_1\right)^2+2\left({ }^{10} C_2\right)^2+3\left({ }^{10} C_3\right)^2+\cdots+10\left({ }^{10} C_{10}\right)^2,$ जहाँ ${ }^{10} C_r, r \in\{1,2, \ldots, 10\}$, द्विपद गुणांकों (binomial coefficients) को दर्शाते हैं। तब $\frac{1}{1430} X$ का मान है ..........|
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
