## .Sam. 3 years ago Guide to Binomial expansion

1. .Sam.

Binomial expansion ============================ nCr, Method Up to “r” terms. Below shows expansion with positive integers Formula goes by $\large ^nC_r(x)^r(n)^{n-r}$ ============================ 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 $(2-7x^2)^6$ 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 $Ans:~64-1344x^2+11760x^4-54880x^6$ -------------------------------------------------- 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 $^8C_r(\frac{6}{x})^r(7x^3)^{8-r}=x^4$ 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, $^8C_r(6x^{-1})^r(7x^3)^{8-r}=x^4$ 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)^{8-r}=x^4 Now you have all the "x" in place, solve it just like normal exponential problems, -r+3(8-r)=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. .Sam.

4) a) Find the terms in x^2 and x^3 in the expansion of (1-3x/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)(1-3x/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+(7x-5x^2))^5 $^5C_0(7x-5x^2)^0(3)^5=243$ $^5C_1(7x-5x^2)^1(3)^4=5(81)(7x-5x^2)=2835x-2025x^2$ $^5C_2(7x-5x^2)^2(3)^3=10(27)(25x^4-70x^3+49x^2)=13230x^2$ 243+2835x-2025x^2+13230x^2 $Ans:~11205x^2+2835x+243$ --------------------------------------------------- 6) Expand and simplify (1+x)^4+(1-x)^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)+(1-4x+6x^2-4x^3+x^4) Add them together, $Ans:~2+12x^2+2x^4$

3. .Sam.

Binomial expansion with negative exponents ------------------------------------------------- Certain rules you have to know: $(b+ax^2)^{-5}$ "b" Must be "1" ------------------------------------------------ Formula: $(1+x)^n=[1+nx+\frac{n(n-1)}{2!}(x^2)+\frac{n(n-1)(n-2)}{3}(x^3)$ Example: 1) Expand the term up to and including x^3 $(7-3x)^{-1}$ 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 $\frac{1}{7}[1+(-1)\frac{-3}{7}x+\frac{(-1)(-1-2)}{2}(\frac{-3}{7}x)^2+\frac{(-1)(-2)(-3)}{6}(\frac{-3}{7}x)^3]$ 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----------- $(3+4x+x^2)[1-\frac{x}{6}+\frac{-\frac{1}{3}(-\frac{1}{3}-1)}{2}(\frac{x^2}{4})]$ $Ans:~\frac{x^4}{18}+\frac{x^3}{18}+\frac{x^2}{2}+\frac{7 x}{2}+3$

4. .Sam.

Done

5. amorfide

very nice, interesting to see it in this method as it is my first time seeing it

6. saifoo.khan

Very comprehensive. COol. ;)

7. .Sam.

@saifoo.khan This should help A2 problems :)

8. saifoo.khan

Negative exponents are a nightmare. D:

9. lgbasallote

@amorfide sam derived that himself...he's cool that way

10. Callisto

Wow~ @.Sam. also joined the tutorial team!!! ._.

11. pratu043

Good way to earn medals.

12. lgbasallote

lol @pratu043 we do these out of the kindness of our hearts ^_^ completely no corruption involved

13. marco26

how about finding the term free of the expansion, if there's any? That would be helpful, too :}

14. pratu043

@lgbasallote no offense meant, I was just voicing my thoughts(anyway, it IS a good way to earn medals isn't it?).

15. pratu043

Must be very tiring to type all that out.

16. Directrix

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?