In the expansion of (1 + x)n, (C1/C0)+2(C2/C1)+3(C3/C2)+ ldots +n (Cn/Cn-1) is equal to

Question

In the expansion of (1 + x)n, (C1/C0)+2(C2/C1)+3(C3/C2)+ ldots +n (Cn/Cn-1) is equal to (A) ((n+1)/2) (B) (n/2) (C) (n(n+1)/2) (D) 3n(n + 1)

Tardigrade Admin · Accepted Answer

Since, we know-that (nCr/nCr-1)=(n-r+1/r) ∴ (C1/C0) = n, (C2/C1) = (n-1/2), (C3/C2)= (n-2/3), ldots Thus, (C1/C0)+2(C2/C1)+3(C3/C2)+ ldots +n (Cn/Cn-1) = n+2⋅ (n-1/1)+3⋅(n-2/3)+ ldots+n⋅(1/n) = [n + (n - 1) + (n - 2) + ... + 1] =(n(n+1)/2)

Q. In the expansion of $(1 + x)^{n}$ ,
$\frac{C _{1}}{C _{0}} + 2 \frac{C _{2}}{C _{1}} + 3 \frac{C _{3}}{C _{2}} + \dots + n \frac{C _{n}}{C _{n - 1}}$ is equal to

$\frac{( n + 1 )}{2}$

$\frac{n}{2}$

$\frac{n ( n + 1 )}{2}$

$3 n (n + 1)$

Q. In the expansion of (1+x)n, C0​C1​​+2C1​C2​​+3C2​C3​​+…+nCn−1​Cn​​ is equal to

2(n+1)​

2n​

2n(n+1)​

3n(n+1)

Since, we know-that nCr−1​nCr​​=rn−r+1​ ∴C0​C1​​=n, C1​C2​​=2n−1​, C2​C3​​=3n−2​, … Thus, C0​C1​​+2C1​C2​​+3C2​C3​​+…+nCn−1​Cn​​ =n+2⋅1n−1​+3⋅3n−2​+…+n⋅n1​ =[n+(n−1)+(n−2)+...+1]=2n(n+1)​

Q. In the expansion of $(1 + x)^{n}$ ,
$\frac{C _{1}}{C _{0}} + 2 \frac{C _{2}}{C _{1}} + 3 \frac{C _{3}}{C _{2}} + \dots + n \frac{C _{n}}{C _{n - 1}}$ is equal to

$\frac{( n + 1 )}{2}$

$\frac{n}{2}$

$\frac{n ( n + 1 )}{2}$

$3 n (n + 1)$