Binomial Expansions: (1 + x)^7; find value of 1.00001^7

[Monkeyseat]

Hi.

Question:

Write down the binomial expansion of (1 + x)^7 in ascending powers of x up to and invluding the term in x^3. Hence determine the value of 1.00001^7 correct to 15 decimal palces.

Working:

(1 + x)^7 = 1^8 + (7C1) * 1^6 * x^1 + (7C2) * 1^5 * x^2 + (7C3) * 1^6 * x^3 + ...

This comes to 1 + 7x + 21x^2 + 35x^3 for up to x^3

I don't have any idea how to do the next part of the question... Could you give me some pointers?

Thanks.

 

[royhaas]

Re: Binomial Expansions

Let \(\displaystyle x=0.00001\).

 

[Monkeyseat]

Re: Binomial Expansions

Let \(\displaystyle x=0.00001\).

Thanks royhass - so is this just a simple substitution?

By comparison if (1 + x)^7 = 1.00001^7, x = 0.00001.

So 1.00001^7 = 1 + 7x + 21x^2 + 35x^3.

Using my calculator to sub x = 0.00001 in:

1 = 1
7x = 0.00007
21x^2 = 0.0000000021
35x^3 = 0.000000000000035.

Added together this gives 1.000070002100035

Is what I have done correct? I can't check this on my calculator as it rounds it after a lot of decimal places.

Thanks for taking the time to reply.