1) (2nCn)2
2) 2nCn
3) 2nCn + 1
4) 2nCn – 1
Solution: (4) 2nCn – 1
\(\begin{array}{l}(1+x)^{n}=n_{C_{0}}+n_{C_{1}} x+n_{C_{2}} x^{2}+\cdots+n_{C_{n}} x^{n} \ and \\ (x+1)^{n}=n_{c_{0}} x^{n}+n_{C_{1}} x^{n-1}+n_{C_{2}} x^{n-2}+\cdots+n_{C_{n}}\\ \text \ On \ multiplying, \\ (1+x)^{n}=\left(n_{C_{0}}+n_{C_{1}} x+n_{C_{2}} x^{2}+\cdots+n_{c_{n}} x^{n}\right) \\ \left(n_{C_{0}} x^{n}+n_{C_{1}} x^{n-1}+n_{C_{2}} x^{n-2}+\cdots+n_{C_{n}}\right)\\ \text \ Comparing \ the \ coefficients \ of \ x^{n} \ on \ both \ sides,\\ 2n_{c_{n}}=\left(n_{C_{0}}\right)^{2}+\left(n_{C_{1}}\right)^{2}+\left(n_{C_{2}}\right)^{2}+\cdots\left(n_{c_{n}}\right)^{2}\\ 2 n_{c_{n}}=1+\left(n_{c_{1}}\right)^{2}+\left(n_{C_{2}}\right)^{2}+\cdots\left(n_{c_{n}}\right)^{2}\\ \Rightarrow\left(n_{c_{1}}\right)^{2}+\left(n_{C_{2}}\right)^{2}+\cdots\left(n_{c_{n}}\right)^{2}\\=2 n_{c_{n}}-1\end{array} \)
