## lilsis76 Group Title how can i find the 100th term in this problem: (1 + x)^100 one year ago one year ago

1. surdawi Group Title

is x the number of terms?

2. lilsis76 Group Title

|dw:1353281221368:dw| this is how the problem is shown. says to find the 100th term. but I never understood :/

3. surdawi Group Title

ok i think that u have to plug 100 for x 101^100 which is lots of numbers

4. lilsis76 Group Title

lol okay so it would be|dw:1353281421656:dw|

5. zepdrix Group Title

Hmm this is a 100th degree binomial. When you expand it out, you should get 101 terms. We want the 100th term that is produced. Are you familiar with pascal's triangle? :D It provides a nice pattern for finding a certain term in binomial expansion.

6. lilsis76 Group Title

yes i know about the pascal triangle.

7. lilsis76 Group Title

but would i really have to write it all out?!

8. zepdrix Group Title

no no no course not :D but we should be able to see a pattern and predict the 100th term hmm

9. zepdrix Group Title

|dw:1353281609783:dw| See how pascal provides the coefficients for each term? With a binomial, they expand in such a way that each term is the product of both terms from the binomial, with the power on the first term starting at 3 (or whatever the degree of the binomial was) and counting down, while the other term starts at 0 and counts up.

10. zepdrix Group Title

There's a formula for generalizing this, but I think that will be alil more confusing than just recognizing the pattern :D

11. lilsis76 Group Title

okay, i see how that works

12. zepdrix Group Title

$\large (1+x)^{100}=$$\large 1^{100}x^0+100*1^{99}x^1+...+100*1^1x^{99}+1^0x^{100}$

13. zepdrix Group Title

There are a bunch of terms in the middle, we wouldn't be able to calculate their coefficients unless we went over the formula for generating them. If we run into a problem like that, we'll have to look at it :D But for now, we care about the 100th term, which is the second to last term.

14. zepdrix Group Title

If you look at pascal's triangle, what can you say about the second AND second to last numbers in any given row?

15. lilsis76 Group Title

well the second number would be the 1, and the second to last number would be i think the 100-1?

16. zepdrix Group Title

Mmm it might be a little confusing because we're actually counting from ZERO, not from 1. So the first row in pascal's triangle corresponds to a binomial of degree zero. So look at the forth row, which corresponds to a binomial of degree 3. The terms are 1 3 3 1 The next row would produce, 1 4 6 4 1. The second and second to last number are 4 yes? :D The 101st row (Which corresponds to a 100th degree binomial) has terms, 1 100 ... 100 1 So the second and second to last numbers are 100 right? :D

17. lilsis76 Group Title

OH!, uh ya i think so cuz, the first set will always be one, and the next number would be the number we are looking for right?

18. zepdrix Group Title

Yes, the coefficient at least :) The coefficient on the second and second to last term will always match the DEGREE of the binomial.

19. zepdrix Group Title

So hmmm.. the 3rd degree binomial (4th row in pascal's triangle) had 4 terms in it. The next row will have 5 terms in it. The 100th degree binomial (101st row in pascal's triangle) will have 101 terms in it. We want the 100th term. So we want the 2nd to last term in that sequence.

20. zepdrix Group Title

These are the last 2 terms in the sequence.. $\large (1+x)^{100}=.....+100(1^1)x^{99}+1(1^0)x^{100}$ Understand where they're coming from? I know this can be a bit confusing :c

21. zepdrix Group Title

More formally you can generalize it like this, $\huge (1+x)^n=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)1^{n-k}x^k$ Where $\large \left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\frac{ n! }{ k!(n-k)! }\right)$

22. zepdrix Group Title

So the 100th term in the sequence would be when n=100, and k=99 (because k starts counting from 0, not from 1). $\large \frac{ 100! }{ 99!(100-99)! }1^{100-99}x^{99}=\frac{100*99!}{99!}1^1x^{99}$ $\huge =100x^{99}$

23. zepdrix Group Title

It's quite a bit to chew on :\ maybe take a look it and try to get through it! :D There is the answer at least. :)

24. lilsis76 Group Title

sorry im back. connection was weird

25. lilsis76 Group Title

okay thank you. an this is super confusing im trying to figure it out still by looking at what you put