We know that <br/>(1+x)n=nC0+nC1x+nC2x2+⋯+nCnxn…… (i) <br/>
and <br/>(x+1)n=nC0xn+nC1xn−1+nC2xn−2+⋯+nCn<br/>
On multiplying equations (i) and (ii), we get <br/><br/>(1+x)2n=(nC0+nC1x+nC2x2+⋯+nCnxn)×<br/>(nC0xn+nC1xn−1+nC2xn−2+⋯+nCn)<br/><br/>
Coefficient of xn in right hand side =(nC0)2+(nC1)2+⋯+(nCn)2
and <br/><br/> coefficient of xn in left hand side =2nCn<br/>∴(nC0)2+(nC1)2+⋯+(nCn)2=n!n!2n!<br/>⇒(nC1)2+⋯+(nCn)2=n!n!(2n)!−1=2nCn−1<br/><br/>