C(n,1)+ 2⋅C(n,2) + ... + (n-1)⋅C(n,n-1) + n⋅C(n,n) = ...?
∑[k=1,n] k⋅C(n,k) =
- AdityaLv 44 years agoFavorite Answer
See binomial equation
[1 + x]^n = C(n,0)+ x*C(n,1) +(x^2)*C(n,2)+ (x^3)*C(n,3) ... + [x^n-1]*C(n,n-1) + (x^n)C(n,n)
Now differentiate this equation on both sides w.r.t. x
n*(1+x)^n-1 = 1*C(n,1)+ (2*x)*C(n,2) + (3x^2)*C(n,3)+ ... + (n-1)*[x^n-2]⋅C(n,n-1) + n*[x^n-1]*C(n,n)
put x = 1, you will see that right part is your question.
n*2^n-1 = C(n,1)+ 2⋅C(n,2) + 3*C(n,3)... + (n-1)⋅C(n,n-1) + n⋅C(n,n)
so Answer = n*[2^n-1]
- NickLv 64 years ago
There are several ways to do this. One way is to note that:
(1+x)^n = ∑[k=0,n] C(n,k)x^k
differentiate both sides wrt x:
n(1+x)^(n-1) = ∑[k=1,n] k⋅C(n+1,k)x^(k-1)
n*2^(n-1) = ∑[k=1,n] k⋅C(n+1,k)
combinatorially we can see that this is the same as counting selections of groups with a leader. The leader is chosen from n individuals then the rest of the group is selected from the n-1 remaining individuals in 2^(n-1) ways. The right hand side chooses the size of the group (k) first then multiplies by the possible leaders of each group (k) then sums over these group sizes.
- 4 years ago
easy Math, 9 is divisible by infinity and it should have been number 1