- #1
JG89
- 728
- 1
I'm just starting to get the hang of Mathematical induction and I was wondering if you guys could please check this proof just to make sure it is correct.
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.
Before I start, I will use (n k) to represent the binomial coefficient.
-----------------
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.