The value of sum limits_{ r =0}^{20}{ }^{50- | vedclass

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Question

$\sum_{r=0}^{20}{ }^{50-r} C_{6}={ }^{50} C_{6}+{ }^{49} C_{6}+{ }^{48} C_{6}+\ldots . .+{ }^{30} C_{6}$

$={ }^{50} C_{6}+{ }^{49} C_{6}+\ldots . .

Vedclass · Accepted Answer

${ }^{51} C _{7}-{ }^{30} C _{7}$

Vedclass · Answer

$\sum_{r=0}^{20}{ }^{50-r} C_{6}={ }^{50} C_{6}+{ }^{49} C_{6}+{ }^{48} C_{6}+\ldots . .+{ }^{30} C_{6}$

$={ }^{50} C_{6}+{ }^{49} C_{6}+\ldots . .+{ }^{31} C_{6}+\left({ }^{30} C_{6}+{ }^{30} C_{7}\right)-{ }^{30} C_{7}$

$={ }^{50} C_{6}+{ }^{49} C_{6}+\ldots . .+\left({ }^{31} C_{6}+{ }^{31} C_{7}\right)-{ }^{30} C_{7}$

$={ }^{50} C_{6}+{ }^{50} C_{7}-{ }^{30} C_{7}$

$={ }^{51} C_{7}-{ }^{30} C_{7}$

${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$

The value of $\sum \limits_{ r =0}^{20}{ }^{50- r } C _{6}$ is equal to

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