The value of { }^{34}C_5 + sum_{i=0}^4 {}^{38 | vedclass

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(B) We use the Pascal's identity: ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$.The given expression is ${}^{34}C_5 + \sum_{i=0}^4 {}^{38-i}C_4 = {}^{34}C_5 +

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${ }^{39}C_5$

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(B) We use the Pascal's identity: ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$.The given expression is ${}^{34}C_5 + \sum_{i=0}^4 {}^{38-i}C_4 = {}^{34}C_5 + {}^{38}C_4 + {}^{37}C_4 + {}^{36}C_4 + {}^{35}C_4 + {}^{34}C_4$.Rearranging the terms,we get $({}^{34}C_5 + {}^{34}C_4) + {}^{35}C_4 + {}^{36}C_4 + {}^{37}C_4 + {}^{38}C_4$.Applying the identity ${}^{34}C_5 + {}^{34}C_4 = {}^{35}C_5$,the expression becomes ${}^{35}C_5 + {}^{35}C_4 + {}^{36}C_4 + {}^{37}C_4 + {}^{38}C_4$.Continuing this process: ${}^{35}C_5 + {}^{35}C_4 = {}^{36}C_5$,then ${}^{36}C_5 + {}^{36}C_4 = {}^{37}C_5$,then ${}^{37}C_5 + {}^{37}C_4 = {}^{38}C_5$,and finally ${}^{38}C_5 + {}^{38}C_4 = {}^{39}C_5$.

The value of ${ }^{34}C_5 + \sum_{i=0}^4 {}^{38-i}C_4$ is

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