Question

The value of 20 sigma 50-r C6 is equal to
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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO CHOOSE THE CORRECT OPTION

$\displaystyle \sf{\sum\limits_{r=0}^{20} \: \large{ {}^{50 - r} C_6}} =$

$1. \: \: \displaystyle \sf{\: \large{ {}^{51} C_7} + \large{ {}^{30} C_7}\: }$

$2. \: \: \displaystyle \sf{\: \large{ {}^{50} C_7} - \large{ {}^{30} C_7}\: }$

$3. \: \: \displaystyle \sf{\: \large{ {}^{51} C_7} - \large{ {}^{30} C_7}\: }$

FORMULA TO BE IMPLEMENTED

$1. \: \: \: \displaystyle \sf{ \large{ {}^{n} C_r} = \:\displaystyle \sf{ \large{ {}^{n} C_{n - r}}} }$

$2. \: \: \: \displaystyle \sf{ \large{ {}^{n} C_r} + \:\displaystyle \sf{ \large{ {}^{n} C_{ r - 1}}} = \large{ {}^{n + 1} C_r} }$

CALCULATION

$\displaystyle \sf{\sum\limits_{r=0}^{20} \: \large{ {}^{50 - r} C_6}}$

$= \displaystyle \sf{\: \large{ {}^{50} C_6} + \large{ {}^{49} C_6}\: + ......... + \large{ {}^{31} C_6} + \large{ {}^{30} C_6}\: }$

$= \displaystyle \sf{\: \large{ {}^{50} C_6} + \large{ {}^{49} C_6}\: + ......... + \large{ {}^{31} C_6} + \large{ {}^{30} C_6} + \large{ {}^{30} C_7} - \large{ {}^{30} C_7}\: }$

$= \displaystyle \sf{\bigg( \large{ {}^{50} C_6} + \large{ {}^{49} C_6}\: + ......... + \large{ {}^{31} C_6} + \large{ {}^{30} C_6} + \large{ {}^{30} C_7} \bigg) - \large{ {}^{30} C_7}\: }$

$= \displaystyle \sf{\bigg( \large{ {}^{50} C_6} + \large{ {}^{49} C_6}\: + ......... + \large{ {}^{31} C_6}+ \large{ {}^{31} C_7} \bigg) - \large{ {}^{30} C_7}\: }$

$= \displaystyle \sf{\bigg( \large{ {}^{50} C_6} + \large{ {}^{49} C_6}\: + ......... + \large{ {}^{32} C_7} \bigg) - \large{ {}^{30} C_7}\: }$

Continuing in the same proces we get

$= \displaystyle \sf{\bigg( \large{ {}^{50} C_6} + \large{ {}^{50} C_7} \bigg) - \large{ {}^{30} C_7}\: }$

$= \displaystyle \sf{ \large{ {}^{51} C_7} - \large{ {}^{30} C_7}\: }$

RESULT

$\boxed{\displaystyle \sf{ \: \: \: \sum\limits_{r=0}^{20} \: \large{ {}^{50 - r} C_6}} \: = \: \large{ {}^{51} C_7} - \large{ {}^{30} C_7}\: \: \: \: }$