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Prove the binomial theorem by induction

Prove the binomial theorem:

Further, prove the formulas:

First, we prove the binomial theorem by induction.
Proof. For the case on the left we have,

On the right,

Hence, the formula is true for the case .
Assume then that the formula is true for some . Then we have,

Thus, if the formula is true for the case then it is true for the case . Hence, we have proved the formula for all

As an application of the binomial theorem we then prove the two formulas.
For the first, apply the binomial theorem with . Then,

For the second, apply the binomial theorem with and . Then,

1. Tiago says:

Nice proof!

Just a tiny observation: actually you could do the base case for , and then you would have proved the binomial theorem for every integer

• Tiago says:

Provided we agree to interpret/define empty sums to equal 0.

2. Marc says:

Thanks, best inductive proof I’ve seen

3. Anonymous says:

Hi,

In line 4, the “a m plus one” should not be taken out, as the first term of the final expression is “b m plus one” only. Taking out the am+1 also means there is an “m-1” term in the summation term, which is currently improperly taken out by reindexing.

• Anonymous says:

^ Correct

4. Darcy says:

In line 3, shouldn’t there be an m choose k in the second summation as well?

• RoRi says:

Yes, in lines 3 and 4. Fixed now. Thanks!