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.

HEY PLEASE HELP ME! :) Find the first four terms in the expansion of (x^1/2+1)^30

Mathematics
See more answers at brainly.com
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
use this : 30C0 (x^1/2)^30(1)^0 + 30C1 (x^1/2)^(30-1) (1)^1 + 30C2 (x^1/2)^(30-2) (1)^2 + 30C3 (x^1/2)^(30-3) (1)^3 calculate and simplify it
wait im confused do you think you can write it out on paper and attach it? im sorry and i understand if you cannot.
do know about combination ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

oh wait! i think i get it! but do i go all the way to thirty? thats whats confusing me?
no...:/
the general formula of combination is |dw:1365545267017:dw| do u know about factorial :)
oh wait! i do know about it loll what would be n and what would be r?
actually, n and r are natural number
so how would i use the formula?
well, first u have to know about factorial defined n! = n*(n-1)*(n-2)*... 3 . 2. 1 an example : 4! = 4 . 3 . 2 . 1 = 24 so far so good ?
yes!
ah, as extra info defined 0! = 1 nah, let's going to combination above 30C0 = 30!/(30-0)!0! = 30!/30!*1 = 30!/30! = 1, right ?
yes!
so, what is the 30C1 ? :) try
30
cool :)
do i go all the way down until i get to 1?
so, so on u can calculation of 30C2 and 30C3 right
yeah
ok, now we have the other problem about exponent
ok
look, that i wrote expression about 1^0, 1^1, 1^2, 1^3. these are be 1 right ? and whatever if multiplied by 1 be its self, right ?
yess
so, u can miss them
then for simplify of (x^1/2)^30, u can use the property of exponent : (a^m)^n = a^(m*n) so, (x^1/2)^30 = x^16, right ?
opp, x^15 i mean :)
yes
then so on, u have to simplify about (x^(1/2))^(30-1) = (x^(1/2))^(29) = x^(29/2) = sqrt(x^29) or u can modif again be sqrt(x^28 * x) = sqrt(x^28) sqrt(x) = (x^(28/2)) * sqrt(x) = x^14 * sqrt(x)
it just simplification in radical form to exponential form
or just one expression sqrt(x^29) only
now, what about u with (x^1/2)^(30-2) (1)^2 = ... ?
ohh!! okay i understand! thank you very much!!!!
you're welcome
:)

Not the answer you are looking for?

Search for more explanations.

Ask your own question