Question

If 
\(\sum\limits_{k=1}^{31}\) \((^{31}C_k) (^{31}C_{k-1})\) \(-\sum\limits_{k=1}^{30}\) \((^{30}C_k) (^{30}C_{k-1})\) \(= \frac{α (60!)} {(30!) (31!)}\)
where \(α ∈ R\), then the value of 16α is equal to

• A. 1411
• B. 1320
• C. 1615
• D. 1855

Answer 1

The Correct Option is A

The correct answer is (A) : 1411
\(\sum\limits_{k=1}^{31}\)\((^{31}C_k) (^{31}C_{k-1})\) \(-\sum\limits_{k=1}^{30}\) \((^{30}C_k) (^{30}C_{k-1})\)
\(=\sum\limits_{k=1}^{31}\) \((^{31}C_k) . (^{31}C_{32-k})\) \(-\sum\limits_{k=1}^{30}\) \((^{30}C_k) . (^{30}C_{k-1})\)
\(= ^{62}C_{32} - ^{60}C_{31}\)
\(= \frac{60!}{31!29!} ( \frac{62.61}{32.30} - 1 ) = \frac{60!}{ 31!29!} \frac{2822}{32.30}\)

\(α = \frac{2822}{32}\)
\(⇒ 16α = 1411\)