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

Answer: (A)

1411

Answer 2

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

$α = \frac{2822}{32}$
$⇒ 16α = 1411$