1. ## Binomial expansion ques

Hi guyz! Can help me with ques 10(ii)?

2. ## Re: Binomial expansion ques

Greetings!

A few notes:

1) Your attached image is upside-down -- most mathematicians don't stand on their heads. As such, please fix this. You will get FAR more (and quicker) help if you present your question neatly. I certainly tend to the neater problems first!

2) Show the board some effort/progress so we know how best to help you. I understand it can be cumbersome to rewrite your work, but you need to put in something to get something. For this problem, I suggest you begin with your steps and solution to the problem's part (i).

Best,
Andy

3. ## Re: Binomial expansion ques

Oh yeah! Sorry for the haphazard posting. I was in a rush just now and thus did not not post the image in an upright manner. Thanks for reminding me of it, abender. Gosh, I didn't even realised it.
But anyway I have got the solution so all is fine and dandy.
But u know I am truly glad to have found this forum where I can indulge in my love of maths with u awesome guys!

4. ## Re: Binomial expansion ques

Originally Posted by Dawncolouredsunlight
Oh yeah! Sorry for the haphazard posting. I was in a rush just now and thus did not not post the image in an upright manner. Thanks for reminding me of it, abender. Gosh, I didn't even realised it.
But anyway I have got the solution so all is fine and dandy.
But u know I am truly glad to have found this forum where I can indulge in my love of maths with u awesome guys!
Excellent job acquiring the solution on your own! Many times when I am stuck on a problem, I will either take a break or do another problem. Revisiting a difficult problem can be enlightening!

5. ## Re: Binomial expansion ques

Haha Yeah man! U know, the thing about maths is that you have to think til your brain hurts to work out the question and having a break is very much needed as it put your brain back in order.
But you know the funny thing is once you get the answer, you want to kick yourself in the ass for not thinking of it earlier!

6. ## Re: Binomial expansion ques

For those who are wondering, and don't want to stand on their heads, the problem is
10 (i) Write down and simplify, in descending powers of x, the first three terms in the expansion of $(x^3+\frac{1}{x^4})^n$, where $n\ge 0$.
The binomial theorem says that $(a+ b)^n= a^n+ na^{n-1}b+ \frac{n(n-1)}{2}a^{n-2}b^2+ \cdot\cdot\cdot+ \frac{n(n-1)}{2}a^2b^{n-2}+ nab^{n-1}+ b^n$. The general "ith term" is $\frac{n!}{i! (n-i)!}a^{n-i}b^i$. In this problem $a= x^3$ and $b= \frac{1}{x^4}= x^{-4}$. The first term is $(x^3)^n= x^{3n}$. The second term is $n(x^3)^{n-1}(x^{-4})= nx^{3n- 7}$. The third term is $\frac{n(n-1)}{2}(x^3)^{n-2}(x^{-4})^2= \frac{n(n-1)}{2}x^{3n- 10}$.

10 (ii) Hence find the value of n given that the coefficient of the third term is 7 times that of the second term.
From the above, that is easy. The second term is $nx^{3x- 7}$ and the third term is $\frac{n(n+1)}{2}x^{3n-10}$. The condition that "the coefficient of the third term is 7 times that of the second term" is that $\frac{n(n+1)}{2}= 7n$. Multiply both sides by 2: $n(n+ 1)= n^2+ n= 14n$. That is equivalent to the quadratic equation $n^2- 13n= n(n- 13)= 0$ n= 0 is impossible here so the answer is n= 13.

10 (iii) Using the value of n found in (ii), without expanding $(x^3+ \frac{1}{x^4})^n$, show that there is no constant term.
As above, the "ith term" of $(a+ b)^n$ is $\frac{n!}{i!(n-i)!}a^{n-i}b^i$. With $a= x^3$ and $b= x^{-4}$, that is $\frac{n!}{i!(n-i)!}(x^3)^{n-i}(x^{-4})^i= \frac{n!}{i!(n-i)!}x^{3n- 7i}$. In particular, with n= 13, as in (ii), that is $\frac{13!}{i!(13- i)!}x^{36-7i}$. That will be a "constant term" if and only if the power of x is 0: 36- 7i= 0. Of course, there is no integer that makes that true.

7. ## Re: Binomial expansion ques

Originally Posted by HallsofIvy
For those who are wondering, and don't want to stand on their heads, the problem is
10 (i) Write down and simplify, in descending powers of x, the first three terms in the expansion of $(x^3+\frac{1}{x^4})^n$, where $n\ge 0$.
The binomial theorem says that $(a+ b)^n= a^n+ na^{n-1}b+ \frac{n(n-1)}{2}a^{n-2}b^2+ \cdot\cdot\cdot+ \frac{n(n-1)}{2}a^2b^{n-2}+ nab^{n-1}+ b^n$. The general "ith term" is $\frac{n!}{i! (n-i)!}a^{n-i}b^i$. In this problem $a= x^3$ and $b= \frac{1}{x^4}= x^{-4}$. The first term is $(x^3)^n= x^{3n}$. The second term is $n(x^3)^{n-1}(x^{-4})= nx^{3n- 7}$. The third term is $\frac{n(n-1)}{2}(x^3)^{n-2}(x^{-4})^2= \frac{n(n-1)}{2}x^{3n- 10}$.
Small typo: the third term is $\dfrac{n(n-1)}{2}x^{3n-14}$.