Binomial theorem (1 Viewer)

[enigma_1]

Evaluate (0.99)^13 using the binomial theorem.. How do you do this?

Thnks

 

[awesome-0_4000]

Try (1-0.01)^13

 

[Squar3root]

im not excatly sure how to do this question but it follows along the same logic as this;

eg: find 49^2

solution:

= 49^2 -1^2

= (49+1)(49-1)

= (50)(48)

= 2400 + 1

= 2401

me thinks you have to use difference of 2 squares somewhere

 

[braintic]

Nup, no difference of two squares. The previous poster has it right.

Were you told how many decimal places / significant figures to include in your answer?
Because each term get progressively smaller, and eventually you'll be able to stop expanding because new terms are insignificant compared to the overall answer.

 

[enigma_1]

Hmm it says correct to 5 sig figs. How would I need to use the Binomial theorem to do this? Oh yep I got the (1-0.01)^13 part but then from there do I actually have to expand it like 5 times? Cause that will take long

 

[braintic]

Hmm it says correct to 5 sig figs. How would I need to use the Binomial theorem to do this? Oh yep I got the (1-0.01)^13 part but then from there do I actually have to expand it like 5 times? Cause that will take long

Not sure what you mean by expand it 5 times. Do you mean take 5 terms? If so, I don't know how many terms I would have to take until I do it.
The first term is 13C0 (1)^13 PLUS
The second is 13C1 (1)^12 (-0.01)^1 PLUS
The third is 13C2 (1)^11 (-0.01)^2 PLUS
The 4th is 13C3 (1)^11 (-0.01)^3

This 4th terms is already only 0.000286.
It won't take many more terms before the answer doesn't change in the 5th significant digit.

 

[enigma_1]

Thanks