sum limits_{k=0}^{6} {}^{51-k}C_{3} is equal | vedclass

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(C) We are given the sum $S = \sum \limits_{k=0}^{6} {}^{51-k}C_{3}$.Expanding the sum,we get $S = {}^{51}C_{3} + {}^{50}C_{3} + {}^{49}C_{3} + {}^{48

Vedclass · Accepted Answer

${}^{52}C_{4}-{}^{45}C_{4}$

Vedclass · Answer

(C) We are given the sum $S = \sum \limits_{k=0}^{6} {}^{51-k}C_{3}$.Expanding the sum,we get $S = {}^{51}C_{3} + {}^{50}C_{3} + {}^{49}C_{3} + {}^{48}C_{3} + {}^{47}C_{3} + {}^{46}C_{3} + {}^{45}C_{3}$.This can be rewritten in ascending order as $S = {}^{45}C_{3} + {}^{46}C_{3} + {}^{47}C_{3} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3}$.Using the Pascal's identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$,we know that ${}^{45}C_{4} + {}^{45}C_{3} = {}^{46}C_{4}$.Adding and subtracting ${}^{45}C_{4}$ to the expression,we get:$S = ({}^{45}C_{4} + {}^{45}C_{3}) + {}^{46}C_{3} + {}^{47}C_{3} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3} - {}^{45}C_{4}$.Using the identity repeatedly:$S = ({}^{46}C_{4} + {}^{46}C_{3}) + {}^{47}C_{3} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3} - {}^{45}C_{4} = {}^{47}C_{4} + {}^{47}C_{3} + \dots + {}^{51}C_{3} - {}^{45}C_{4}$.Continuing this process,we eventually get ${}^{51}C_{4} + {}^{51}C_{3} - {}^{45}C_{4} = {}^{52}C_{4} - {}^{45}C_{4}$.

$\sum \limits_{k=0}^{6} {}^{51-k}C_{3}$ is equal to

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