Finding Power an Expression is Raised to, when Coefficient of x^3 is Known

[cmkluza]

Hello, I'm having some trouble trying to figure out this problem. The statement is:

When \(\displaystyle (1 + \frac{x}{2})^n, n \in N \), is expanded in ascending powers of \(\displaystyle x \), the coefficient of \(\displaystyle x^3 \) is 70.

(a) Find the value of \(\displaystyle n \).

I've tried doing some expansions of the equation at different values of \(\displaystyle n \), up to 5, but I've been unable to figure out the pattern. For some of the terms I've been able to figure out what the term will be in relation to \(\displaystyle x \) and \(\displaystyle n \), such as \(\displaystyle \frac{n}{2}x \) , but I can't see any pattern for \(\displaystyle x^3 \) yet. Any suggestions?

Thanks for any help!

 

[Deleted member 4993]

Hello, I'm having a little bit of trouble with the following problem:

When \(\displaystyle (1 + \frac{x}{2})^n, n \in N \), is expanded in ascending powers of \(\displaystyle x \), the coefficient of \(\displaystyle x^3 \) is 70.

(a) Find the value of \(\displaystyle n \).

I've tried doing out the problem to a few different values of \(\displaystyle n \), up to 5, but haven't been able to see much. I found a pattern such that the \(\displaystyle x \) coefficient is equal to [\frac{n}{2}x [/tex], but have been unable to find anything like that to do with \(\displaystyle x^3 \). Any ideas?

Thanks for any help!

Do you know the formula for binomial expansion - expansion of (1 + x)^m?

If not, do a google search and come back to tell us what information you found.

 

[pka]

When \(\displaystyle (1 + \frac{x}{2})^n, n \in N \), is expanded in ascending powers of \(\displaystyle x \), the coefficient of \(\displaystyle x^3 \) is 70.
(a) Find the value of \(\displaystyle n \).

Can you simplify \(\displaystyle \dbinom{n}{3}\left(\dfrac{x}{2}\right)^3~?\)

 

[cmkluza]

Do you know the formula for binomial expansion - expansion of (1 + x)^m?

If not, do a google search and come back to tell us what information you found.

Thank you very much for your help! I apologize for not replying to my own thread sooner, but I've been very busy. I can't believe that I haven't seen binomial expansion before now, as it seems like a very basic procedure that I just never learned. It makes me wonder what else I'm missing. I believe I was able to properly solve the problem with this help, after getting it down to a cubic and doing some algebra.

Again, I appreciate your assistance!

 

[cmkluza]

Can you simplify \(\displaystyle \dbinom{n}{3}\left(\dfrac{x}{2}\right)^3~?\)

Thanks for your help! I hadn't learned much about binomial expansion prior to having read about it here, a flaw on my part I suppose, but now I can understand that equation, and its significance. I appreciate your help!