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Mathematics Question on Binomial theorem

The value of the sum ( nC1)2+( nC2)2+( nC3)2+...+( nCn)2(\,^nC_1)^2+(\,^nC_2)^2+(\,^nC_3)^2+...+(\,^nC_n)^2 is

A

( 2nCn)2(\,^{2n}C_n)^2

B

 2nCn\,^{2n}C_n

C

 2nCn+1\,^{2n}C_n+1

D

 2nCn−1\,^{2n}C_n-1

Answer

 2nCn−1\,^{2n}C_n-1

Explanation

Solution

We know that (1+x)n=nC0+nC1x+nC2x2+⋯+nCnxn…… (i) (1+ x )^{ n }={ }^{ n } C _{0}+{ }^{ n } C _{1} x +{ }^{ n } C _{2} x ^{2}+\cdots+{ }^{ n } C _{ n } x ^{ n } \quad \ldots \ldots \text { (i) } and (x+1)n=nC0xn+nC1xn−1+nC2xn−2+⋯+nCn(x+1)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}+{ }^{n} C_{2} x^{n-2}+\cdots+{ }^{n} C_{n} On multiplying equations (i) and (ii), we get (1+x)2n=(nC0+nC1x+nC2x2+⋯+nCnxn)× (nC0xn+nC1xn−1+nC2xn−2+⋯+nCn)\begin{array}{l} (1+x)^{2 n}=\left({ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\cdots+{ }^{n} C_{n} x^{n}\right) \times \\\ \left({ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}+{ }^{n} C_{2} x^{n-2}+\cdots+{ }^{n} C_{n}\right) \end{array} Coefficient of xnx ^{ n } in right hand side =(nC0)2+(nC1)2+⋯+(nCn)2=\left({ }^{ n } C _{0}\right)^{2}+\left({ }^{ n } C _{1}\right)^{2}+\cdots+\left({ }^{ n } C _{ n }\right)^{2} and  coefficient of xn in left hand side =2nCn ∴(nC0)2+(nC1)2+⋯+(nCn)2=2n!n!n! ⇒(nC1)2+⋯+(nCn)2=(2n)!n!n!−1=2nCn−1\begin{array}{l} \text { coefficient of } x ^{ n } \text { in left hand side }={ }^{2 n } C _{ n } \\\ \therefore\left({ }^{ n } C _{0}\right)^{2}+\left({ }^{ n } C _{1}\right)^{2}+\cdots+\left({ }^{ n } C _{ n }\right)^{2}=\frac{2 n !}{ n ! n !} \\\ \Rightarrow\left({ }^{ n } C _{1}\right)^{2}+\cdots+\left({ }^{ n } C _{ n }\right)^{2}=\frac{(2 n ) !}{ n ! n !}-1={ }^{2 n } C _{ n }-1 \end{array}