# Binomial Theorem - Questions

• Nov 24th 2008, 04:01 PM
cnmath16
Binomial Theorem - Questions
32. For the expansion of (k+t)^22, state:

a) the number of terms (1 mark)

b) the degree of each term (1 mark)

c) the first four terms in the expansion, without coefficients (2 marks)

33. Find the first five terms in the expansion of (r-2s)^8. (5 marks)

34. Find the term not involving 'y' in the expansion of (1÷y^3 + y^2)^5. (4 marks)

Any help on any of the above questions would be greatly, greatly appreciated. They are questions from a Grade 12 Data Management course that I am learning on my own.
• Nov 26th 2008, 07:46 AM
JD-Styles
32 a) (x=y)^n has n+1 terms. So (k+t)^22 has 23 terms.

b) Every term in the expansion is of the form $\displaystyle k^{22-i}t^i$ (this is ignoring the coefficients). So their degree is (22-i)+i=22.

c) If we ignore coefficients, we can use what we said in part b) with i=0, 1, 2, 3. This gives

$\displaystyle k^{22},\, k^{21}t,\, k^{20}t^2,\, k^{19}t^3$

33. Using the binomial expansion formula:

$\displaystyle \sum_{i=0}^8 \begin{pmatrix} 8 \\ i \end{pmatrix} r^{8-i}(-2s)^i$

So, the first five terms would be $\displaystyle r^8, \, -16r^7s, \, 112r^6s^2, \, -448r^5s^3, \, and \,\,\,1120r^4s^4$.

34. Ignoring the coefficients, every term will be of the form $\displaystyle y^{-3(5-i)}y^{2i}=y^{5i-15}$. So, if we want the term without y, we need the exponant to be zero, thus 5i-15=0 or i=3. Now all we have to do is calculate the coefficent for i=3, which is simply $\displaystyle \begin{pmatrix} 5 \\ 3 \end{pmatrix}=10$.

Hope that helps.
• Nov 26th 2008, 09:02 AM
cnmath16
Thanks a lot! I was able to figure out most of them on my own, but your input definitely helped put it all together.

I just have one more question that I was wondering if you could guide me through..

Find the term involving 'y' in the expansion of (1÷ y^3 + y^2)^5
• Nov 26th 2008, 11:19 AM
JD-Styles
That one seems strange, because there is no term with 'y'. Because if we follow the same logic as in 34, we need the exponant to be 1, thus 5i-15=1 or i=16/5, which isn't an integer. So either the question is wrong or they just want you to say 0.
• Nov 26th 2008, 03:57 PM
cnmath16
Is it at all possible for you to show your work for questions 33 and 34, so that I can get an idea of how my between stuff compares to yours? (I set up my equation a bit differently)

Thank you!!
• Nov 27th 2008, 07:35 PM
JD-Styles
Sure.

33. Using the binomial expansion formula $\displaystyle \sum_{i=0}^8 \begin{pmatrix} 8 \\ i \end{pmatrix} r^{8-i}(-2s)^i$, the first term is i=0, the second is i=1, and so on.

1st term (i=0): $\displaystyle \begin{pmatrix} 8 \\ 0 \end{pmatrix} r^{8-0}(-2s)^0=(1)r^8(1)=r^8$

2nd term (i=1): $\displaystyle \begin{pmatrix} 8 \\ 1 \end{pmatrix} r^{8-1}(-2s)^1=(8)r^7(-2s)=-16r^7s$

3rd term (i=2): $\displaystyle \begin{pmatrix} 8 \\ 2 \end{pmatrix} r^{8-2}(-2s)^2=(28)r^6(4s^2)=112r^6s^2$

4th term (i=3): $\displaystyle \begin{pmatrix} 8 \\ 3 \end{pmatrix} r^{8-3}(-2s)^3=(56)r^5(-8s^3)=-448r^5s^3$

5th term (i=4): $\displaystyle \begin{pmatrix} 8 \\ 4 \end{pmatrix} r^{8-4}(-2s)^4=(70)r^4(16s^4)=1120r^4s^4$

34. The binomial expansion formula gives us $\displaystyle \sum_{i=0}^5 \begin{pmatrix} 5 \\ i \end{pmatrix} (1/y^3)^{5-i}(y^2)^i=\sum_{i=0}^5 \begin{pmatrix} 5 \\ i \end{pmatrix} y^{-3(5-i)}y^{2i}=\sum_{i=0}^5 \begin{pmatrix} 5 \\ i \end{pmatrix} y^{-15+3i+2i}=\sum_{i=0}^5 \begin{pmatrix} 5 \\ i \end{pmatrix} y^{-15+5i}$

I showed how I got that i=3 is the term we're looking for (-15+5i=0 or i=3), so let's evaluate the term i=3 (which is the 4th term):

$\displaystyle \begin{pmatrix} 5 \\ 3 \end{pmatrix} y^{-15+5(3)}=(10) y^{-15+15}=10y^0=10(1)=10$

Hope that's detailed enough (Happy)