Question

Find each coefficient described.
1) Coefficient of x^2 in expansion of (2 + x)^5

Answer 1

I know that I must use the binomial theorem in this

Answer 2

but I don't know where to start. Would I compress the expression (2+x)^5 until the second row in Pascal's triangle?

Answer 3

Like plug in all the numbers and stuff?

Answer 4

\[T _{r+1}=C _{r}^{5}2^{5-r}x^r\]
to get x^2 put r=2
and solve to get the reqd. solution.

Answer 5

So everywhere that there is an r, I plug in a 2?

Answer 6

reqd. co-efficient is \[C _{2}^{5}2^{5-2}=?\]

Answer 7

um.. that's a bit confusing. Could I show you how I would do it?

Answer 8

I would use the 6th row in Pascal's triangle, and plug the first term (degree-0) and second term (0-degree) in it

Answer 9

I would do this because it is being raised to the 5th power

Answer 10

\[=\frac{ 5*4 }{ 2*1 }*2^3=?\]
show your work.

Answer 11

well yeah, I would do

1x^5 2^0

and so on

Answer 12

with the numbers 1, 5, 10, 10, 5, 1 from the 6th row in pascal's triangle

Answer 13

i solved it taking general term

Answer 14

what do you mean?

Answer 15

general term in the expansion of \[\left( x+a \right)^n~is~T _{r+1}=C _{r}^{n}x ^{n-r}a^r\]

Answer 16

ooh.. I have never seen that before? Is that the Binomeal Theorem?

Answer 17

*Bionomial

Answer 18

it is general term in the binomial expansion of (x+a)^n

Answer 19

Well, I have never seen that. I was watching a few Khan Academy videos, but I still don't get it. How do you use that formula?

Answer 20

\[C _{r}^{n}=\frac{ n*(n-1)*(n-2)...(n-(r-1)) }{ r*(r-1)*(r-2)*...*1 }\]

Answer 21

\[C _{r}^{n}=\frac{ n*(n-1)*(n-2)*...(n-(r-1))*(n-r)! }{ r*(r-1)(r-2)*...1*(n-r)! }=\frac{ n! }{ r!*\left( n-r)! \right) }\]

Answer 22

thank you for all the help!! But I figured it out!!

Answer 23

yw