Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

how can i find the 100th term in this problem: (1 + x)^100

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

is x the number of terms?
|dw:1353281221368:dw| this is how the problem is shown. says to find the 100th term. but I never understood :/
ok i think that u have to plug 100 for x 101^100 which is lots of numbers

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

lol okay so it would be|dw:1353281421656:dw|
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.
yes i know about the pascal triangle.
but would i really have to write it all out?!
no no no course not :D but we should be able to see a pattern and predict the 100th term hmm
|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.
There's a formula for generalizing this, but I think that will be alil more confusing than just recognizing the pattern :D
okay, i see how that works
\[\large (1+x)^{100}=\]\[\large 1^{100}x^0+100*1^{99}x^1+...+100*1^1x^{99}+1^0x^{100}\]
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.
If you look at pascal's triangle, what can you say about the second AND second to last numbers in any given row?
well the second number would be the 1, and the second to last number would be i think the 100-1?
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
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?
Yes, the coefficient at least :) The coefficient on the second and second to last term will always match the DEGREE of the binomial.
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.
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
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 \[\large \left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\frac{ n! }{ k!(n-k)! }\right)\]
So the 100th term in the sequence would be when n=100, and k=99 (because k starts counting from 0, not from 1). \[\large \frac{ 100! }{ 99!(100-99)! }1^{100-99}x^{99}=\frac{100*99!}{99!}1^1x^{99}\] \[\huge =100x^{99}\]
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. :)
sorry im back. connection was weird
okay thank you. an this is super confusing im trying to figure it out still by looking at what you put

Not the answer you are looking for?

Search for more explanations.

Ask your own question