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.Sam.Best 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 
 one year ago

.Sam.Best 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, \[Ans:~2+12x^2+2x^4\]
 one year ago

.Sam.Best 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 \[(3+4x+x^2)[1\frac{x}{6}+\frac{\frac{1}{3}(\frac{1}{3}1)}{2}(\frac{x^2}{4})]\] \[Ans:~\frac{x^4}{18}+\frac{x^3}{18}+\frac{x^2}{2}+\frac{7 x}{2}+3\]
 one year ago

amorfideBest 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
 one year ago

saifoo.khanBest ResponseYou've already chosen the best response.1
Very comprehensive. COol. ;)
 one year ago

.Sam.Best ResponseYou've already chosen the best response.19
@saifoo.khan This should help A2 problems :)
 one year ago

saifoo.khanBest ResponseYou've already chosen the best response.1
Negative exponents are a nightmare. D:
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
@amorfide sam derived that himself...he's cool that way
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
Wow~ @.Sam. also joined the tutorial team!!! ._.
 one year ago

pratu043Best ResponseYou've already chosen the best response.0
Good way to earn medals.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
lol @pratu043 we do these out of the kindness of our hearts ^_^ completely no corruption involved
 one year ago

marco26Best 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 :}
 one year ago

pratu043Best 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?).
 one year ago

pratu043Best ResponseYou've already chosen the best response.0
Must be very tiring to type all that out.
 one year ago

DirectrixBest 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?
 one year ago
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