Question

If ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+\cdot \cdot \cdot \cdot \cdot +C_n{x}^n$, then

• A. $\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+\cdot \cdot \cdot \cdot +n\frac{C_n}{C_{n-1}}=\frac{n(n+1)}{2}$
• B. $C_{n-1}=n$
• C. $\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+\cdot \cdot \cdot \cdot +n\frac{C_n}{C_{n-1}}=\frac{n(n-1)}{2}$
• D. None of the above

Answer 1

Answer: (A), (B)

$C_{n-1}=n$
$\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+\cdot \cdot \cdot \cdot +n\frac{C_n}{C_{n-1}}=\frac{n(n+1)}{2}$

$\frac{{}^n{\text{C}}_1}{{}^n{\text{C}}_0}+2\frac{{}^n{\text{C}}_2}{{}^n{\text{C}}_1}+3\frac{{}^n{\text{C}}_3}{{}^n{\text{C}}_2}+...$

$=\frac{n}{1}+\frac{2(n-1)}{2}+\frac{3(n-2)}{3}+...$

$=n+(n-1)+(n-2)+...+1$

$=n.n-(1+2+....+(n-1\left)\right)$

$=\frac{n(n+1)}{2}$

Also,
${}^n{\text{C}}_{n-1}=\frac{n!}{(n-1)!1!}=n$

Hence, options A and B are the correct answers.