Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# The value of (nC1)2 + (nC2)2 + (nC3)2 +... + (nCn)2 is

1) (2nCn)2

2) 2nCn

3) 2nCn + 1

4) 2nCn – 1

Solution: (4) 2nCn – 1

$$\begin{array}{l}(1+x)^{n}=n_{C_{0}}+n_{C_{1}} x+n_{C_{2}} x^{2}+\cdots+n_{C_{n}} x^{n} \ and \\ (x+1)^{n}=n_{c_{0}} x^{n}+n_{C_{1}} x^{n-1}+n_{C_{2}} x^{n-2}+\cdots+n_{C_{n}}\\ \text \ On \ multiplying, \\ (1+x)^{n}=\left(n_{C_{0}}+n_{C_{1}} x+n_{C_{2}} x^{2}+\cdots+n_{c_{n}} x^{n}\right) \\ \left(n_{C_{0}} x^{n}+n_{C_{1}} x^{n-1}+n_{C_{2}} x^{n-2}+\cdots+n_{C_{n}}\right)\\ \text \ Comparing \ the \ coefficients \ of \ x^{n} \ on \ both \ sides,\\ 2n_{c_{n}}=\left(n_{C_{0}}\right)^{2}+\left(n_{C_{1}}\right)^{2}+\left(n_{C_{2}}\right)^{2}+\cdots\left(n_{c_{n}}\right)^{2}\\ 2 n_{c_{n}}=1+\left(n_{c_{1}}\right)^{2}+\left(n_{C_{2}}\right)^{2}+\cdots\left(n_{c_{n}}\right)^{2}\\ \Rightarrow\left(n_{c_{1}}\right)^{2}+\left(n_{C_{2}}\right)^{2}+\cdots\left(n_{c_{n}}\right)^{2}\\=2 n_{c_{n}}-1\end{array}$$