# Proof using the binomial theorem

1. Mar 1, 2012

### naaa00

1. The problem statement, all variables and given/known data
Use the binomial theorem to rpove that for n a positive integer we have:

(1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! product(r=0 to k-1) (1 - r/n)]

3. The attempt at a solution

(1 + 1/n)^n = 1 + sum(k=1 to n) (n choose r) 1^n-k (1/n)^k, where (n choose r) = n!/r!(n - r)!, the binomial coefficients.

I'm trying to fit "n!/r!(n - r)!" to an expression that involves the products, since product(k=1 to n) n = n!

The product on the RHS I rewrite it as: product(r=0 to k-1) [(n - r)/n]

[product(r=0 to k-1) (n - r)] x [product(r=0 to k-1) (1/n)]

=> (n - r)! (1/n)!

So...

(1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! (n-r)! (1/n)!] or (1 + 1/n)^n = 1 + sum(k=1 to n) [(n-r)!/k!n!]

I don't like this. I feel that all of this is taking me nowhere.

Any ideas will be very appreciated.

2. Mar 1, 2012

### issacnewton

Hi

look at the snapshot i have attached. in the second equation, its clear that we can express
it as the term on RHS in your expression.

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3. Mar 1, 2012

### Joffan

Each of the terms of the RHS sum corresponds to one of the expanded terms of the LHS.