How do you find the coefficient of #x# in the expansion of #(x+3)^5#?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

4
Euan S. Share
Jul 19, 2016

Answer:

Coefficient is 405.

Explanation:

We use the binomial theorem:

#(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k)y^k#

where #((n),(k)) = (n!)/(k!(n-k)!)#

For #n=5# and #y=3# we are looking for the coefficient of #x^1#. This means we need #n-k = 1 implies k = 4#.

#((5),(4))x^1*3^4 = (5!)/(4!1!)*81*x = 405x#

Was this helpful? Let the contributor know!
1500