Let X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2) | vedclass

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(D) The given expression is $X = \sum_{r=1}^{10} r \left({ }^{10} C_r\right)^2$.Using the property ${ }^{n} C_r = { }^{n} C_{n-r}$,we have $X = \sum_{

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(D) The given expression is $X = \sum_{r=1}^{10} r \left({ }^{10} C_right)^2$.Using the property ${ }^{n} C_r = { }^{n} C_{n-r}$,we have $X = \sum_{r=1}^{10} (10-r) \left({ }^{10} C_{10-r}ight)^2 = \sum_{r=0}^{9} (10-r) \left({ }^{10} C_right)^2$.Adding the two forms of $X$:$2X = \sum_{r=1}^{10} r \left({ }^{10} C_right)^2 + \sum_{r=0}^{9} (10-r) \left({ }^{10} C_right)^2 = 10 \left({ }^{10} C_0ight)^2 + \sum_{r=1}^{9} 10 \left({ }^{10} C_right)^2 + 10 \left({ }^{10} C_{10}ight)^2 = 10 \sum_{r=0}^{10} \left({ }^{10} C_right)^2$.Using the identity $\sum_{r=0}^{n} ({ }^{n} C_r)^2 = { }^{2n} C_n$,we get $2X = 10 \cdot { }^{20} C_{10}$.Thus,$X = 5 \cdot { }^{20} C_{10}$.Given ${ }^{20} C_{10} = 184756$,we have $X = 5 	imes 184756 = 923780$.Finally,$\frac{X}{1430} = \frac{923780}{1430} = 646$.

Let $X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2)^2 + 3({ }^{10} C _3)^2 + \ldots + 10({ }^{10} C _{10})^2$,where ${ }^{10} C _{ r }$ for $r \in \{1, 2, \ldots, 10\}$ denotes binomial coefficients. Then,the value of $\frac{1}{1430} X$ is:

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