# Prove the equality 1+1/2*C(n,1)+c(n,2)/3+..+C(n,n)/(n+1)=[2^(n+1)-1]/(n+1) .

*print*Print*list*Cite

### 1 Answer

Supposing that C(n,k) are binomial coefficients, we'll recall the binomial theorem:

(1 + x)^n = (1 + C(n,1)*x + C(n,2)*x^2 + ... + C(n,k)*x^k + ... + C(n,n)*x^n)

If we'll consider the terms of binomial theorem as functions andÂ if we'll determine the definite integral of the terms of binomial theorem, we'll get:

Int (1+x)^ndx = Int [1 + C(n,1)*x + C(n,2)*x^2 + ... + C(n,k)*x^k + ... + C(n,n)*x^n]dx

We'll impose the limits of integration x = 0 to x = 1

We'll manage the left side:

Int (1+x)^ndx = (1 + x)^(n+1)/(n+1) F(1)

We'll apply Leibniz Newton:

F(1) = (1 + 1)^(n+1)/(n+1) = (2)^(n+1)/(n+1)

F(0) = (1 + 0)^(n+1)/(n+1) = (1)^(n+1)/(n+1)

Int (1+x)^ndx = F(1) - F(0)

Int (1+x)^ndx = (2)^(n+1)/(n+1) - (1)^(n+1)/(n+1)

We'll factorize and we'll get:

Int (1+x)^ndx = [(2)^(n+1) - 1]/(n+1) (1)

We'll determine now the definite integral of each term of binomial expanssion:

Int [1 + C(n,1)*x + C(n,2)*x^2 + ... + C(n,k)*x^k + ... + C(n,n)*x^n]dx = Int dx + Int C(n,1)*x dx + .. + Int C(n,n)*x^n]dx

Int [1 + C(n,1)*x + C(n,2)*x^2 + ... + C(n,k)*x^k + ... + C(n,n)*x^n]dx = x + C(n,1)*x^2/2 + .... + C(n,n)*x^(n+1)/(n+1)

We'll apply Leibniz Newton:

F(1) = 1 + C(n,1)*1^2/2 + .... + C(n,n)*1^(n+1)/(n+1)

F(0) = 0 + C(n,1)*0^2/2 + .... + C(n,n)*0^(n+1)/(n+1)

Int [1 + C(n,1)*x + C(n,2)*x^2 + ... + C(n,k)*x^k + ... + C(n,n)*x^n]dx = F(1) - F(0)

Int [1 + C(n,1)*x + C(n,2)*x^2 + ... + C(n,k)*x^k + ... + C(n,n)*x^n]dx = 1 + C(n,1)/2 + .... + C(n,n)/(n+1) (2)

Comparing (1) and (2), we notice that are exactly the terms from the equality that has to be demonstrated.

**Based on binomial theorem, the requested inequality 1+1/2*C(n,1)+c(n,2)/3+..+C(n,n)/(n+1)=[2^(n+1)-1]/(n+1) is verified, for any natural number n.**