Question

Coefficient!

Answer 1

\[\LARGE (1+x^{2})^{40}(x^{2}+2+\frac{1}{x^{2}})\]
co-eff of x^20=?

Answer 2

\[\LARGE x^{10}(1+x)^{30}\]
im stuck here..

Answer 3

@waterineyes

Answer 4

What are you doing....trying to do?

Answer 5

finding co-eff of x^20?

Answer 6

oh, so supposing you did expand it, but without doing it, you wanna find the coefficient hmmm

Answer 7

What is the original expression?

Answer 8

read the 1st reply

Answer 9

It helps to know the binomial theorem.
\[\huge (a+b)^n=\sum_{k=0}^n \binom{n}{k}a^{k} b^{n-k}\]

Answer 10

So well say a = 1, b = x... so our coefficient for the 20th term is gonna be \(\binom{40}{20}\)

Answer 11

But such isn't taking into account some the other factor.

Answer 12

oops, also b = x^2

Answer 13

\((1 + x^{2})^{40}\) has a term with \(x^{18}\;and\;x^{20}\;and\;x^{22}\). Find those three and then worry about the second factor.

Answer 14

So the first thing I would do is just say: \[ \Large
x^{20} = (x^{2})^{10}
\]Then we need to find the 10th term in the binomial \((1+x^2)^{40}\). It's going to be just \(\binom{40}{10}\)

Answer 15

...and then the other two.

Answer 16

For the second term \(x^2 + 2 + \frac{1}{x^2}\) what we could read this as is, multiply it by 2 then add the previous and following term coefficients.

Answer 17

Co-efficient of \[\LARGE x^{20}\] in x^{10} (1+x^2)^30
then its written
Co-efficient of x^10 in (1+x^2)^30
how

Answer 18

Do you understand the binomial theorem?

Answer 19

yes

Answer 20

Do you understand what I've explained?

Answer 21

kind of

Answer 22

okay got it..thanks