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Before I start, I will use (n k) to represent the binomial coefficient.

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Prove 2^n = 1 + (n 1) + (n 2) + ... + (n n-1) + (n n)

I am first going to find the case for 2^(n+1).

Using the formula they just gave for 2^n, I will multiply both sides by 2. So,

2^(n+1) = 2[1 + (n 1) + (n 2) + ... + (n n-1) + (n n)]

I will call that equation, equation 1.

Now, using the binomial theorem, I will simplify it for (2 + 0) ^ (n+1)

(2 + 0)^(n+1) = (n+1 0)2^(n+1) + (n+1 1)(2^n(0)

We can stop right there since b = 0 , and we can say

(2 + 0)^(n+1) = (n+1 0)2^(n+1)

I will call that equation, equation 2.

Now I am going to equate equation 1 and equation 2 and see if they are the same.

2[1 + (n 1) + (n 2) + ... + (n n-1) + (n n)] = (n+1 0)2^(n+1)

2[1 + (n 1) + (n 2) + ... + (n n-1) + (n n)] = 2(2^n)(n+1 0)

1 + (n 1) + (n 2) + ... + (n n-1) + (n n) = 2^n(n+1 0)

1 + (n 1) + (n 2) + ... + (n n-1) + (n n) = 2^n * [(n+1)!/(n+1)!]

1 + (n 1) + (n 2) + ... + (n n-1) + (n n) = 2^n

And that was the original formula that I was suppose to prove. So, now I check the case for 2^1 with the original formula. It works out to 2.

Therefore, the theorem is correct.