Question

$c_0,c_1,c_2$ denotes coefficents expansion of $(1+x{)}^n$ , then $c_1+c_1{c}_2+c_2{c}_3+.......c_{n-1}{c}_n=\frac{\left(2n\right)!}{(n+1)!(n-1)!}$

• A. True
• B. False

Answer 1

The correct option is A True

$C_0,C_1,C_2$ denotes coefficients expansion of $(1+x{)}^{\text{n}}$ then

$C_1+C_1{C}_2+C_2{C}_3+\dots +C_{n-1}{C}_n=\frac{2n!}{(n-1)!(n+1)!}$

we know that

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\cdots +C_{n-2}{x}^{n-2}$

$+C_{n-1}{x}^{n-1}+C_n{x}^n$

Also we can write it as

$(1+x{)}^n=C_n{x}^n+C_{n-1}{x}^{n-1}+\cdots C_2{x}^2+C_{1}x+C_0$

Multiplying these two expressions

$(1+x{)}^{2n}=(C_0+C_{1}x+C_2{x}^2+\dots +C_{n-1}{x}^{n-1}+C_n{x}^n)X$

$(C_n{x}^n+C_{n-1}{x}^{n-1}+\dots +C_2{x}^2+C_{1}x+C_0)$

Now equate the coefficients of

$x^{n-r}$ from both sides of equations

We get

${}^{2n}{C}_{n-r}=C_0{C}_{n-r}+C_1{C}_{n-r-1}+C_2{C}_{n-r-2}+C_3{C}_{n-r-3}+\dots +C_{n-r}{C}_n$

Put $r=1$

$C_0{C}_1+C_1{C}_2+C_2{C}_3+\cdots +C_{n-1}{C}_n=\frac{2n!}{(n-1)!(n+1)!}$