Question

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

• A. \({}^{39}C_4\)
• B. \({}^{39}C_5\)
• C. \({}^{38}C_5\)
• D. \({}^{38}C_4\)

Hint:

The identity \[ {}^nC_r+{}^nC_{r+1}={}^{n+1}C_{r+1} \] is very useful for simplifying sums of consecutive binomial coefficients.

Answer 1

The Correct Option is B

Step 1: Expand the summation.
Given expression is \[ {}^{34}C_5+\sum_{i=0}^{4}{}^{38-i}C_4 \] Now, \[ \sum_{i=0}^{4}{}^{38-i}C_4 = {}^{38}C_4+{}^{37}C_4+{}^{36}C_4+{}^{35}C_4+{}^{34}C_4 \] Therefore, the expression becomes \[ {}^{34}C_5+{}^{38}C_4+{}^{37}C_4+{}^{36}C_4+{}^{35}C_4+{}^{34}C_4 \]

Step 2: Apply hockey-stick identity.
Using the identity \[ {}^nC_r+{}^nC_{r+1}={}^{n+1}C_{r+1}, \] we first combine \[ {}^{34}C_4+{}^{34}C_5={}^{35}C_5 \] So the expression becomes \[ {}^{38}C_4+{}^{37}C_4+{}^{36}C_4+{}^{35}C_4+{}^{35}C_5 \] Again, \[ {}^{35}C_4+{}^{35}C_5={}^{36}C_5 \] Now the expression becomes \[ {}^{38}C_4+{}^{37}C_4+{}^{36}C_4+{}^{36}C_5 \] Again, \[ {}^{36}C_4+{}^{36}C_5={}^{37}C_5 \] Now the expression becomes \[ {}^{38}C_4+{}^{37}C_4+{}^{37}C_5 \] Again, \[ {}^{37}C_4+{}^{37}C_5={}^{38}C_5 \] Now the expression becomes \[ {}^{38}C_4+{}^{38}C_5 \] Finally, \[ {}^{38}C_4+{}^{38}C_5={}^{39}C_5 \]

Step 3: Final conclusion.
Therefore, \[ \boxed{{}^{39}C_5} \]