Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

.Sam. Group TitleBest ResponseYou've already chosen the best response.19
Binomial expansion ============================ nCr, Method Up to “r” terms. Below shows expansion with positive integers Formula goes by \[\large ^nC_r(x)^r(n)^{nr}\] ============================ Lets say: 1) Expand fully \[ (3+2x)^5\] Using the formula, \[ ^5C_0(2x)^0(3)^5=243\] \[ ^5C_1(2x)^1(3)^4=5(2x)(81)=810x\] \[ ^5C_2(2x)^2(3)^3=10(4x^2)(27)=1080x^2\] \[ ^5C_3(2x)^3(3)^2=10(8x^3)(9)=720x^3\] \[ ^5C_4(2x)^4(3)^1=5(16x^4)(3)=240x^4\] \[^5C_5(2x)^5(3)^0=32x^5\] \[Ans:~ 243+810x+1080x^2+720x^3+240x^4+32x^5\]  2) Expand up to and including x^6 term for \[(27x^2)^6\] Using the formula, \[ ^6C_0(7x^2)^0(2)^6=64\] \[ ^6C_1(7x^2)^1(2)^5=6(7x^2)(32)=1344x^2\] \[ ^6C_2(7x^2)^2(2)^4=15(49x^4)(16)=11760x^4\] \[ ^6C_3(7x^2)^3(2)^3=20(343x^6)(8)=54880x^6\] Notice you stop expanding at x^6 \[Ans:~641344x^2+11760x^454880x^6\]  3) Find the coefficient of x^4 in the expansion of \[ (7x^3+\frac{6}{x})^8\] Now they only wants the coefficient of x^4, so its tedious to expand all the terms and then find the coefficient right? Theres a way to find,so, by using back the formula \[ ^8C_r(\frac{6}{x})^r(7x^3)^{8r}=x^4\] You basically don't know what's "r" so you'll have to use the expansion equals to x^4 (Now, here you'll need some good exponential skills) We set the 6/x to 6x^1 to avoid confusion, \[ ^8C_r(6x^{1})^r(7x^3)^{8r}=x^4\] From here, you just "ignore" all the coefficents first, so you just ignore the nCr, by doing that, you'll get (x^{1})^r(x^3)^{8r}=x^4 Now you have all the "x" in place, solve it just like normal exponential problems, r+3(8r)=4 r=5 Now you have the "r" Substitute into the formula and solve \[ ^8C_5(\frac{6}{x})^5(7x^3)^{3}=56(\frac{7776}{x^5})(343x^9)=149361408x^4\] Ans: 149361408 NOTE: When they ask find the term independent of x, means you find the constant, Eg. x^0 
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.19
4) a) Find the terms in x^2 and x^3 in the expansion of (13x/2)^6 Basically for this you should know that 6C2 and 6C3 will give you x^2 and x^3 So, Using the formula, \[ ^6C_2(3x/2)^2(1)^4=15(\frac{9x^2}{4})=\frac{135}{4}x^2\] \[ ^6C_3(3x/2)^3(1)^3=20(\frac{27}{8}x^3)=\frac{135}{2}x^3\] \[\frac{135}{4}x^2\frac{135}{2}x^3\] b)Given that there is no term in x^3 in the expansion of (k+2x)(13x/2)^6, find the value of the constant k. (\frac{270}{4}x^3\frac{135}{2}x^3) By comparing both terms, k=1  5)Expanding quadratics, Expand up to and including x^2 term for (3+(7x5x^2))^5 \[ ^5C_0(7x5x^2)^0(3)^5=243\] \[ ^5C_1(7x5x^2)^1(3)^4=5(81)(7x5x^2)=2835x2025x^2\] \[ ^5C_2(7x5x^2)^2(3)^3=10(27)(25x^470x^3+49x^2)=13230x^2\] 243+2835x2025x^2+13230x^2 \[Ans:~11205x^2+2835x+243\]  6) Expand and simplify (1+x)^4+(1x)^4 The trick here's that you don't expand both the binomial, you just do one of it its enough. When you expand (1+x)^4, you should get 1+4x+6x^2+4x^3+x^4 Now,when I have (x)^1=x (x)^2=x^2 (x)^3=x^3 You can see there's a pattern here, Even powers end up positive, odd number ends up negative. o, (1+4x+6x^2+4x^3+x^4)+(14x+6x^24x^3+x^4) Add them together, \[Ans:~2+12x^2+2x^4\]
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.19
Binomial expansion with negative exponents  Certain rules you have to know: \[(b+ax^2)^{5}\] "b" Must be "1"  Formula: \[(1+x)^n=[1+nx+\frac{n(n1)}{2!}(x^2)+\frac{n(n1)(n2)}{3}(x^3)\] Example: 1) Expand the term up to and including x^3 \[(73x)^{1}\] First you have to make the "b" into 1, The rule here is take out the 7 from the brackets, then tag it with the power from the brackets and make all other terms divide by that number, \[7^{1}(1\frac{3x}{7})^{1}\] by setting n=1, x=3x/7 Then expand \[\frac{1}{7}[1+(1)\frac{3}{7}x+\frac{(1)(12)}{2}(\frac{3}{7}x)^2+\frac{(1)(2)(3)}{6}(\frac{3}{7}x)^3]\] Simplifying you'll get \[Ans:~ \frac{1}{7}+\frac{3 x}{49}+\frac{9 x^2}{343}+\frac{27 x^3}{2401}\]  Expand up to first 3 terms \[\frac{3+4x+x^2}{\sqrt[3]{1+\frac{x}{2}}}\] Rearranging \[(3+4x+x^2)(1+\frac{x}{2})^{1/3}\] n=1/3 , x=x/2 Expand \[(3+4x+x^2)[1\frac{x}{6}+\frac{\frac{1}{3}(\frac{1}{3}1)}{2}(\frac{x^2}{4})]\] \[Ans:~\frac{x^4}{18}+\frac{x^3}{18}+\frac{x^2}{2}+\frac{7 x}{2}+3\]
 2 years ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
very nice, interesting to see it in this method as it is my first time seeing it
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.1
Very comprehensive. COol. ;)
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.19
@saifoo.khan This should help A2 problems :)
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.1
Negative exponents are a nightmare. D:
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
@amorfide sam derived that himself...he's cool that way
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.0
Wow~ @.Sam. also joined the tutorial team!!! ._.
 2 years ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
Good way to earn medals.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
lol @pratu043 we do these out of the kindness of our hearts ^_^ completely no corruption involved
 2 years ago

marco26 Group TitleBest ResponseYou've already chosen the best response.0
how about finding the term free of the expansion, if there's any? That would be helpful, too :}
 2 years ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
@lgbasallote no offense meant, I was just voicing my thoughts(anyway, it IS a good way to earn medals isn't it?).
 2 years ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
Must be very tiring to type all that out.
 2 years ago

Directrix Group TitleBest ResponseYou've already chosen the best response.0
I recall learning that the sum of the coefficients of the expansion of (x + y)^n = (1 + 1) ^n = 2 The sum of the coefficients of the expansion of (3 + 2x)^5 = (3 + 2)^5 = 5^5 = 3125. I checked that sum against your example. Does anyone think that the sum of the coefficients of the expansion of (3x  4) ^ 30 would be (3  4) ^ 30 = (1)^30 = 1?
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.