If displaystyle∑k=131( 31 Ck)( 31 C k -1)- displaystyle ∑ k =130( 30 C k )( 30 C k -1)=(α(60 !)/(30 !)(31 !)), where α ∈ R, then the value of 16 α is equal to

Question

If displaystyle∑k=131( 31 Ck)( 31 C k -1)- displaystyle ∑ k =130( 30 C k )( 30 C k -1)=(α(60 !)/(30 !)(31 !)), where α ∈ R, then the value of 16 α is equal to (A) 1411 (B) 1320 (C) 1615 (D) 1855

Tardigrade Admin · Accepted Answer

displaystyle∑t=131 31 C t ⋅ 31 C t -1 = 31 C 1 ⋅ 31 C 0+ 31 C 2 ⋅ 31 C 1+ ldots .+ 31 C 31 ⋅ 31 C 30 = 31 C 0 ⋅ 31 C 30+ 31 C 1 ⋅ 31 C 29+ ldots . .++ 31 C 30 ⋅ 31 C 0= 62 C 30 ⋅ Similarly displaystyle ∑ R =130( 30 C R ⋅ 30 C R -1)= 60 C 29 62 C 30= 62 C 29=(62 !/30 ! 32 !)-(60 !/29 ! 31 !)=(60 !/29 ! 31 !) (62.61/30.32)-1 (60 !/30 ! 31 !)((2822/32)) ∴ 16 α=16 X (2822/32)=1411

Q. If k=1∑31​(31Ck​)(31Ck−1​)−k=1∑30​(30Ck​)(30Ck−1​)=(30!)(31!)α(60!)​, where α∈R, then the value of 16α is equal to

1411

1320

1615

1855

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