The value of sum_{r=0}^{20} {}^{50-r}C_{6} is | vedclass

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(B) We need to evaluate the sum $S = \sum_{r=0}^{20} {}^{50-r}C_{6} = {}^{50}C_{6} + {}^{49}C_{6} + \dots + {}^{30}C_{6}$.Using the identity ${}^{n}C_

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${}^{51}C_{7} - {}^{30}C_{7}$

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(B) We need to evaluate the sum $S = \sum_{r=0}^{20} {}^{50-r}C_{6} = {}^{50}C_{6} + {}^{49}C_{6} + \dots + {}^{30}C_{6}$.Using the identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$,we can rewrite the sum.Note that ${}^{30}C_{6} = {}^{31}C_{7} - {}^{30}C_{7}$.Thus,$S = {}^{50}C_{6} + {}^{49}C_{6} + \dots + {}^{31}C_{6} + ({}^{31}C_{7} - {}^{30}C_{7})$.Using the identity repeatedly: ${}^{n}C_{r} + {}^{n+1}C_{r+1} = {}^{n+1}C_{r+1}$,we get:${}^{31}C_{6} + {}^{31}C_{7} = {}^{32}C_{7}$.Continuing this process,the sum telescopes to ${}^{51}C_{7} - {}^{30}C_{7}$.

The value of $\sum_{r=0}^{20} {}^{50-r}C_{6}$ is equal to

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