Even our teacher can't do this! "The 3rd term of (1+x)^n is 36x^2. Find 4th term...."

Leah5467

New member
Joined
Feb 28, 2019
Messages
41
Binomial stuff:

a The third term of \(\displaystyle (1\, +\, x)^n\) is \(\displaystyle 36x^2.\) Find the fourth term.

b If \(\displaystyle (1\, +\, kx)^n\, =\, 1\, -\, 12x\, +\, 60x^2\, -\, ...,\) find the values of \(\displaystyle k\) and \(\displaystyle n\).

c Find \(\displaystyle a\) if the coefficient of \(\displaystyle x^{11}\) in the expansion of \(\displaystyle \left(x^2\, +\, \frac{1}{ax}\right)^{10}\) is 15.

I am stuck from the beginning. I tried using general term way:Tr+1,and gives me the equation of (n chooses 2)=36,and i don't know how to solve it
 

Attachments

Last edited by a moderator:

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,595
Did you intend to include a picture of the problem? Please do so, and also type what you can of the problem, and show whatever work you are able to do.

Clearly you have not read our posting guidelines; to do so, start here. We are not here to do the work for you; we work with you to help you learn, and to do that, we need to know what you know and where you are stuck.
 

Leah5467

New member
Joined
Feb 28, 2019
Messages
41
I am stuck from the beginning. I tried using general term way:Tr+1,and gives me the equation of (n chooses 2)=36,and i don't know how to solve it
 

Harry_the_cat

Senior Member
Joined
Mar 16, 2016
Messages
1,238
Binomial stuff:

a The third term of \(\displaystyle (1\, +\, x)^n\) is \(\displaystyle 36x^2.\) Find the fourth term.

b If \(\displaystyle (1\, +\, kx)^n\, =\, 1\, -\, 12x\, +\, 60x^2\, -\, ...,\) find the values of \(\displaystyle k\) and \(\displaystyle n\).

c Find \(\displaystyle a\) if the coefficient of \(\displaystyle x^{11}\) in the expansion of \(\displaystyle \left(x^2\, +\, \frac{1}{ax}\right)^{10}\) is 15.

I am stuck from the beginning. I tried using general term way:Tr+1,and gives me the equation of (n chooses 2)=36,and i don't know how to solve it
Use the formula \(\displaystyle {n\choose r} = \frac{n!}{r! (n-r)!} \)

and the fact that \(\displaystyle \frac {n!}{(n-2)!} = n(n-1)\)
 
Last edited by a moderator:

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,335
For the second one, start with

\(\displaystyle (-\ 1)^1 * \dbinom{n}{1} * 1^{(n-1)} * (kx)^1 = -\ 12x \)

and

\(\displaystyle (-\ 1)^2 * \dbinom{n}{2} * 1^{(n-2)} * (kx)^2 = 60x^2.\)

This simplifies to a system of two linear equations in two unknowns.
 
Top