Question

If $(2\le r\le n)$, then ${{}^{n}C}_r+2{{}^{n}C}_{r+1}+{{}^{n}C}_{r+2}$ is equal to

• A. $2.{{}^{n+2}C}_{r+2}$
• B. ${{}^{n+r}C}_{r+1}$
• C. ${{}^{n+2}C}_{r+2}$
• D. ${{}^{n+1}C}_r$

Answer 1

Answer: (C)

${{}^{n+2}C}_{r+2}$

To solve this, we'll use the relation ${}^n{C}_r{+}^n{C}_{r+1}{=}^{n+1}{C}_{r+1}$ ...[1]

${}^n{C}_r+2^n{C}_{r+1}{+}^n{C}_{r+2}$

$={(}^n{C}_r{+}^n{C}_{r+1})+{(}^n{C}_{r+1}{+}^n{C}_{r+2})$

${=}^{n+1}{C}_{r+1}{+}^{n+1}{C}_{r+2}$ ...(Using [1])

${=}^{n+2}{C}_{r+2}$ ...(Using [1])