The value of _r=0^20^50-rC_6 is equal to:

Question

The value of $\displaystyle\sum_{r=0}^{20}{}^{50-r}C_6$ is equal to:

Accepted Answer

$$\sum\limits_{r = 0}^{20} {} {}^{50 - r}{C_6} = {}^{50}{C_6} + {}^{49}{C_6} + {}^{48}{C_6} + .... + {}^{30}{C_6}$$

$$ = {}^{50}{C_6} + {}^{49}{C_6} + .... + {}^{31}{C_6} + ({}^{30}{C_6} + {}^{30}{C_7}) - {}^{30}{C_7}$$

$$ = {}^{50}{C_6} + {}^{49}{C_6} + .... + ({}^{31}{C_6} + {}^{31}{C_7}) - {}^{30}{C_7}$$

$$ = {}^{50}{C_6} + {}^{50}{C_7} - {}^{30}{C_7}$$

$= {}^{51}{C_7} - {}^{30}{C_7}$

[As $${{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}}$$]

Solutions · 1