# Understanding The Binomial Theorem for (a+b)^n: I've been given Thrm for (1+b)^n...?

#### Ted_Grendy

##### New member
Hi all

I was wondering if someone could clear up some doubt in my mind.

I am trying to get my head around the Binomial Theorem.

The Binomial Theorem states:-

(a+b)^n = a^n +na^n-1b + n(n-1)/2! * a^n-2*b^2 + n(n-1)(n-2)/3! * a^n-3*b^3 + ...

The problem is that I have been given a Binomial Theorem which is different to the above, it states:-

(1+b)^n = 1 + nb + n(n-1) * b^2/2! + n(n-1)(n-2)b^3/3! + ....

Are they the same where does the 1 come from in the (1+b) side? shouldn't it be a ???

Can someone help?

Thank you.

#### topsquark

##### Full Member
Hi all

I was wondering if someone could clear up some doubt in my mind.

I am trying to get my head around the Binomial Theorem.

The Binomial Theorem states:-

(a+b)^n = a^n +na^n-1b + n(n-1)/2! * a^n-2*b^2 + n(n-1)(n-2)/3! * a^n-3*b^3 + ...

The problem is that I have been given a Binomial Theorem which is different to the above, it states:-

(1+b)^n = 1 + nb + n(n-1) * b^2/2! + n(n-1)(n-2)b^3/3! + ....

Are they the same where does the 1 come from in the (1+b) side? shouldn't it be a ???

Can someone help?

Thank you.
They are essentially the same expansion. It's the difference between (a + b) and (1 + b). We simply sub in a = 1.

-Dan

#### Jomo

##### Elite Member
Hi all

I was wondering if someone could clear up some doubt in my mind.

I am trying to get my head around the Binomial Theorem.

The Binomial Theorem states:-

(a+b)^n = a^n +na^n-1b + n(n-1)/2! * a^n-2*b^2 + n(n-1)(n-2)/3! * a^n-3*b^3 + ...

The problem is that I have been given a Binomial Theorem which is different to the above, it states:-

(1+b)^n = 1 + nb + n(n-1) * b^2/2! + n(n-1)(n-2)b^3/3! + ....

Are they the same where does the 1 come from in the (1+b) side? shouldn't it be a ???

Can someone help?

Thank you.
It is the same formula except a is 1. So it can not be used in all cases like the other formula. As a result I would not memorize this formula since it is only valid for a=1.

It reminds me of when teachers push the FOIL method down students throats. It is ridiculous for a student to learn a special formula for (a+b)(c+d) when they have to learn another method for multiplying other than 2 terms times 2 terms. The latter method also works for the former. What are these teachers and authors doing???!!!!!

#### Dr.Peterson

##### Elite Member
The Binomial Theorem states:-

(a+b)^n = a^n +na^n-1b + n(n-1)/2! * a^n-2*b^2 + n(n-1)(n-2)/3! * a^n-3*b^3 + ...

The problem is that I have been given a Binomial Theorem which is different to the above, it states:-

(1+b)^n = 1 + nb + n(n-1) * b^2/2! + n(n-1)(n-2)b^3/3! + ....

Are they the same where does the 1 come from in the (1+b) side? shouldn't it be a ???
As has been said, the second is just a special case of the first, taking a=1.

But you can also reconstruct the first if all you know is the second, simpler form. If you know it as

(1+r)^n = 1 + nr + n(n-1) * r^2/2! + n(n-1)(n-2)r^3/3! + ...

and you want to find (a+b)^n, just rewrite this as a^n * (1 + b/a)^n, and replace r with (b/a):

(1+(b/a))^n = 1 + n(b/a) + n(n-1) * (b/a)^2/2! + n(n-1)(n-2)(b/a)^3/3! + ...

(a+b)^n = a^n * (1 + b/a)^n = 1 + n a^n (b/a) + n(n-1) a^n (b/a)^2/2! + n(n-1)(n-2) a^n (b/a)^3/3! + ...

and so on. So although it's probably easier to just remember the long form, you don't lose anything in the long run if you learn the special case instead. They are equivalent.

#### JeffM

##### Elite Member
Hi all

I was wondering if someone could clear up some doubt in my mind.

I am trying to get my head around the Binomial Theorem.

The Binomial Theorem states:-

(a+b)^n = a^n +na^n-1b + n(n-1)/2! * a^n-2*b^2 + n(n-1)(n-2)/3! * a^n-3*b^3 + ...

The problem is that I have been given a Binomial Theorem which is different to the above, it states:-

(1+b)^n = 1 + nb + n(n-1) * b^2/2! + n(n-1)(n-2)b^3/3! + ....

Are they the same where does the 1 come from in the (1+b) side? shouldn't it be a ???

Can someone help?

Thank you.
Do you know sigma notation?

$$\displaystyle \displaystyle \sum_{j=k}^n f(j) = f(k) + f(k + 1) + ... f(n - 1) + f(n).$$

It basically says add everything up varying j by 1 and starting with j = k and ending with j = n.

Any problem? So in sigma notation the binomial theorem is

$$\displaystyle n \in \mathbb Z^+ \implies (a + b)^n = \displaystyle \sum_{j=0}^n \dbinom{n}{j} * a^{(n-j)} * b^j.$$

Now let us apply that formula when b = 1.

$$\displaystyle n \in \mathbb Z^+ \implies (a + 1)^n = \displaystyle \sum_{j=0}^n \dbinom{n}{j} * a^{(n-j)} * 1^j = \sum_{j=0}^n \dbinom{n}{j} a^{(n-j)} * 1 = \sum_{j=0}^n \dbinom{n}{j} a^{(j-n)}.$$

In other words, we have a special case of a general formula as was said in many previous posts.