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

#### Leah5467

##### New member
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

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#### Dr.Peterson

##### Elite Member
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
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

##### Full Member
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 $$\displaystyle \frac {n!}{(n-2)!} = n(n-1)$$

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#### JeffM

##### Elite Member
$$\displaystyle (-\ 1)^1 * \dbinom{n}{1} * 1^{(n-1)} * (kx)^1 = -\ 12x$$
$$\displaystyle (-\ 1)^2 * \dbinom{n}{2} * 1^{(n-2)} * (kx)^2 = 60x^2.$$