Question

how(nc0)2+ (nc1)2 + (nc2)2 + .......+ (ncn)2 =2ncn

Answer 1

Answer:

nco=1

nc1=n

nc2=nc2n[n_1]/2

---------------

nc2=nc2n[n_1]/2

ncn=cn

now

nco=1

nc1=2n

ncn=cn

=2ncn

Answer 2

The given result can be proved by using a bit of manipulation.

• The left hand side , nc0² + nC1² + nC2² +.......+ nCn² can also be written as nC0.nCn + nC1.nCn-1 + nC2.nCn-2 + ........ + nCn.nC0 (since nCr=nCn-r).
• We can notice that the sum of subscipts of the two elements being multiplied in each term of the sum is equal to n.
• So, the above expression can be expressed as the coefficient of x^n in (1+x)^n.(1+x)^n.
• Which makes it the coefficient of x^n in (1+x)^2n , hence it is equal to 2nCn.