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If #( 1 + x )^n = C_0 + C_1 x_1 + C_2 x_2 + ⋯ + C_n x_n# then show that #C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)=((2n)!)/((n+r)!(n-r)!) ?#

1 Answer
Feb 3, 2018

Answer:

Please see below.

Explanation:

We know that

#(a+b)^n=C_0^na^n+C_1^na^(n-1)b+C_2^na^(n-2)b^2+ ⋯ + C_n^nb^n#

and #r^(th)# term is #C_r^na^rb^(n-r)# and #C_r^n=(n!)/(r!(n-r)!)#

and hence #(1+x)^n=C_0 1^n+C_1 1^(n-1)*x+C_2 1^(n-2)x^2+ ⋯ + C_nx^n#

= #(1+x)^n=C_0 1^n+C_1x+C_2x^2+ ⋯ + C_nx^n# (A)

and similarly #(x+1)^n=C_0 x^n+C_1 x^(n-1)*1+C_2 x^(n-2)1^2+ ⋯ + C_n1^n#

= #(x+1)^n=C_0x^n+C_1x^(n-1)+C_2x^(n-2)+ ⋯ +C_n1# (B)

Multipllying (A) and (B), we get #(1+x)^n(x+1)^n# on the LHS and multiplication of expansion on the RHS.

Now what is coefficient of #x^(n+r)# in this.

While on LHS we have #(1+x)^(2n)# and coefficient of #x^(n+r)# is #(2n!)/((n+r)!(2n-n-r)!)=(2n!)/((n+r)!(n-r)!)#

On the RHS we get #x^(n+r)#, when #C_0x^n# is multiplied by #C_rx^r#; #C_1x^(n-1)# is multiplied by #C_(r+1)x^(r+1)#; #C_2x^(n-2)# is multiplied by #C_(r+2)x^(r+2)# and so on till we get #C_nx^n# multipled by #C_(r+n)x^(n-(r+n)#.

Hence coefficient of #x^(n+r)# is

#C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)#

and hence #C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)=((2n)!)/((n+r)!(n-r)!)#