JEE MAIN - Mathematics (2003 - No. 44)
If $${}^n{C_r}$$ denotes the number of combination of n things taken r at a time, then the expression $$\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$$ equals
$$\,{}^{n + 1}{C_{r + 1}}$$
$${}^{n + 2}{C_r}$$
$${}^{n + 2}{C_{r + 1}}$$
$$\,{}^{n + 1}{C_r}$$
คำอธิบาย
Arrange it this way,
$$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$$
$$ = {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$$
$$\left[ \, \right.$$ Now use the rule,
$$\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]$$
$$ = {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr$$
$$ = {}^{n + 2}{C_{r + 1}}$$
$$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$$
$$ = {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$$
$$\left[ \, \right.$$ Now use the rule,
$$\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]$$
$$ = {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr$$
$$ = {}^{n + 2}{C_{r + 1}}$$