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.
The expression is $ {}^{34}C_5 + \sum_{i=0}^{4} {}^{38-i}C_4 $. Expanding the sum: $ {}^{38}C_4 + {}^{37}C_4 + {}^{36}C_4 + {}^{35}C_4 + {}^{34}C_4 $. So the full expression is: $ {}^{34}C_5 + {}^{38}C_4 + {}^{37}C_4 + {}^{36}C_4 + {}^{35}C_4 + {}^{34}C_4 $.
Step 2: Recall Pascal's identity.
Pascal's identity says $ {}^n C_r + {}^n C_{r+1} = {}^{n+1} C_{r+1} $. We will use this repeatedly, starting from the smallest terms.
Step 3: Combine the two terms with n=34.
$ {}^{34}C_4 + {}^{34}C_5 = {}^{35}C_5 $. The expression becomes: $ {}^{38}C_4 + {}^{37}C_4 + {}^{36}C_4 + {}^{35}C_4 + {}^{35}C_5 $.
Step 4: Combine the two terms with n=35.
$ {}^{35}C_4 + {}^{35}C_5 = {}^{36}C_5 $. The expression becomes: $ {}^{38}C_4 + {}^{37}C_4 + {}^{36}C_4 + {}^{36}C_5 $.
Step 5: Keep applying Pascal's identity.
$ {}^{36}C_4 + {}^{36}C_5 = {}^{37}C_5 $, giving $ {}^{38}C_4 + {}^{37}C_4 + {}^{37}C_5 $. Then $ {}^{37}C_4 + {}^{37}C_5 = {}^{38}C_5 $, giving $ {}^{38}C_4 + {}^{38}C_5 $. Finally $ {}^{38}C_4 + {}^{38}C_5 = {}^{39}C_5 $.
Step 6: State the final answer.
\[ \boxed{{}^{39}C_5} \]