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

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(B) The given sum is $S = \sum_{r=0}^{20} {}^{50-r}C_{6} = {}^{50}C_{6} + {}^{49}C_{6} + {}^{48}C_{6} + \dots + {}^{30}C_{6}$.Using the Hockey-stick i

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

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(B) The given sum is $S = \sum_{r=0}^{20} {}^{50-r}C_{6} = {}^{50}C_{6} + {}^{49}C_{6} + {}^{48}C_{6} + \dots + {}^{30}C_{6}$.Using the Hockey-stick identity,which states that $\sum_{i=r}^{n} {}^{i}C_{r} = {}^{n+1}C_{r+1}$,we can rewrite the sum.First,note that ${}^{30}C_{6} = {}^{31}C_{7} - {}^{30}C_{7}$.Alternatively,we can use the property ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$ repeatedly:$S = {}^{50}C_{6} + {}^{49}C_{6} + \dots + {}^{31}C_{6} + {}^{30}C_{6}$.Adding and subtracting ${}^{30}C_{7}$:$S = ({}^{50}C_{6} + {}^{49}C_{6} + \dots + {}^{30}C_{6} + {}^{30}C_{7}) - {}^{30}C_{7}$.Using the identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$ iteratively:${}^{30}C_{6} + {}^{30}C_{7} = {}^{31}C_{7}$.${}^{31}C_{6} + {}^{31}C_{7} = {}^{32}C_{7}$.Continuing this process up to ${}^{50}C_{6} + {}^{50}C_{7} = {}^{51}C_{7}$.Thus,the sum simplifies 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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