Question

If $C_0,C_1,C_2,\cdots C_n$, are the binomial coefficients of the expansion $(1+x{)}^n$, where $n$ is even, then $C_0+(C_0+C_1)+(C_0+C_1+C_2)+\cdots \cdots +(C_0+C_1+C_2+\cdots +C_{n-1})=$

• A. $n\cdot 2^{n-1}$
• B. $n\cdot 2^{n-1}-n$
• C. $n\cdot 2^n+n$
• D. $(n-1)\cdot 2^n$

Answer 1

The correct option is A $n\cdot 2^{n-1}$

Given expansion is $(1+x{)}^n$
Let
$S=C_0+(C_0+C_1)+(C_0+C_1+C_2)+\cdots \cdots +(C_0+C_1+C_2+\cdots +C_{n-1})\Rightarrow S=n\cdot C_0+(n-1)C_1+(n-2)C_2+\cdots +C_{n-1}\Rightarrow S=\sum _{r=0}^{n-1}(n-r){}^n{C}_r\Rightarrow S=\sum _{r=0}^{n-1}n{}^n{C}_r-\sum _{r=0}^{n-1}r{}^n{C}_r\Rightarrow S=n[2^n-{}^n{C}_n]-n\sum _{r=1}^{n-1}{}^{n-1}{C}_{r-1}\Rightarrow S=n[2^n-{}^n{C}_n]-n[2^{n-1}-1]\therefore S=n\cdot 2^{n-1}$