Guide to Binomial expansion

Question

Tutorials · Answer

$\style{padding:5pt}{\huge \text{Binomial expansion}} $

nCr, Method
Up to “r” terms.

1) Here we have an expansion with positive integers, $ \mathbb{Z} $
$ \color{teal}{\large ^nC_r(x)^r(n)^{n-r}} $

$\text{For example:}$

1) Expand fully 
$ (3+2x)^5$

Using the formula,
$^5C_0(2x)^0(3)^5=243$

$^5C_1(2x)^1(3)^4=5(2x)(81)=810x$

$^5C_2(2x)^2(3)^3=10(4x^2)(27)=1080x^2$

$^5C_3(2x)^3(3)^2=10(8x^3)(9)=720x^3$

$^5C_4(2x)^4(3)^1=5(16x^4)(3)=240x^4$

$^5C_5(2x)^5(3)^0=32x^5$

$Answer:~ 243+810x+1080x^2+720x^3+240x^4+32x^5$

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2) Expand up to and including $x^6$ term for 
$(2-7x^2)^6$

Using the formula, $^nC_r(x)^r(n)^{n-r}$

$^6C_0(-7x^2)^0(2)^6=64$

$^6C_1(-7x^2)^1(2)^5=6(-7x^2)(32)=-1344x^2$

$^6C_2(-7x^2)^2(2)^4=15(49x^4)(16)=11760x^4$

$^6C_3(-7x^2)^3(2)^3=20(-343x^6)(8)=-54880x^6$

Now, notice you have to stop expanding at $x^6$

$Answer:~64-1344x^2+11760x^4-54880x^6$

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3) Find the coefficient of $x^4$ in the expansion of

$(7x^3+\frac{6}{x})^8$

Now all you need to find is the coefficient of $x^4$, notice that it's tedious to expand all the terms and then find the coefficient
There is another way to do it, by revisiting the formula  $^nC_r(x)^r(n)^{n-r}$

We get
$^8C_r(\frac{6}{x})^r(7x^3)^{8-r}=x^4$

Basically we don't know what "r" is so what you can do is to use the expansion equals to x^4 (Now, here you'll need some good exponential skills)
We're gonna set the $\frac{6}{x}$ to $6x^{-1}$ to avoid confusion,

$^8C_r(6x^{-1})^r(7x^3)^{8-r}=x^4$

Now, we're gonna ignore all the coefficients for a moment, so we just ignore the nCr, by doing that, we will get
$(x^{-1})^r(x^3)^{8-r}=x^4$
Now that we have all the "x" in place, we can solve it easily
$-r+3(8-r)=4$
$r=5$
Now that we have the "r"
Substitute into the formula and solve
$^8C_5(\frac{6}{x})^5(7x^3)^{3}=56(\frac{7776}{x^5})(343x^9)=149361408x^4$

$Answer: 149361408$

NOTE: When they ask find the term independent of x, it means you've got to find the constant, Eg. $x^0$

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(4)
$\mathbb{\text{A) Find the terms in $x^2$ and $x^3$ in the expansion of $(1-\frac{3x}{2})^6$}}$

In this problem you should know that 6C2 and 6C3 gives you $x^2$ and $x^3$
Back to our formula, $^nC_r(x)^r(n)^{n-r}$

$^6C_2(-\frac{3x}{2})^2(1)^4=15(\frac{9x^2}{4})=\frac{135}{4}x^2$

$^6C_3(-\frac{3x}{2})^3(1)^3=20(\frac{-27}{8}x^3)=-\frac{135}{2}x^3$

$\frac{135}{4}x^2-\frac{135}{2}x^3$

$ \mathbb{\text{B)Given that there is no term in $x^3$ in the expansion of $(k+2x)(1-\frac{3x}{2})^6$,}}$
$\mathbb{\text{find the value of the constant k.}}$

$(\frac{270}{4}x^3-\frac{135}{2}x^3)$
By comparing both terms, k=1

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$\Large \mathbb{\text{(5) Expanding quadratics}}$

Expand up to and including $x^2$ term for $(3+(7x-5x^2))^5$
$^5C_0(7x-5x^2)^0(3)^5=243 \ \ ^5C_1(7x-5x^2)^1(3)^4=5(81)(7x-5x^2)=2835x-2025x^2 \ \ ^5C_2(7x-5x^2)^2(3)^3=10(27)(25x^4-70x^3+49x^2)=13230x^2 \ \ 243+2835x-2025x^2+13230x^2$

$Answer:~11205x^2+2835x+243$

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$\Large \mathbb {\text{(6) Expand and simplify}}$

$(1+x)^4+(1-x)^4$

The catch here's that you don't expand both the binomial, you just do one of it is 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 a pattern here, Even powers end up positive, odd number ends up negative.
$(1+4x+6x^2+4x^3+x^4)+(1-4x+6x^2-4x^3+x^4)$

Add them together,
$Ans: 2+12x^2+2x^4$

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$\Large \mathbb {\text{Binomial expansion with negative exponents}}$

There are Certain rules you will need to know: $(b+ax^2)^{-5}$ "b" Must be "1"

Formula: $(1+x)^n=[1+nx+\frac{n(n-1)}{2!}(x^2)+\frac{n(n-1)(n-2)}{3}(x^3)$ For example: $\text{1) Expand the term up to and including $x^3$}$ $(7-3x)^{-1}$ First you have to make the "b" equals 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 $n=-1, x=\frac{-3x}{7}$ Then we 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 $Answer:~ \frac{1}{7}+\frac{3 x}{49}+\frac{9 x^2}{343}+\frac{27 x^3}{2401}$

Another example, expand up to first 3 terms $\frac{3+4x+x^2}{\sqrt[3]{1+\frac{x}{2}}}$ Rearranging it $(3+4x+x^2)(1+\frac{x}{2})^{-1/3}$ $n=-\frac{1}{3} , x=\frac{x}{2}$ Expanding it we get $(3+4x+x^2)[1-\frac{x}{6}+\frac{-\frac{1}{3}(-\frac{1}{3}-1)}{2}(\frac{x^2}{4})$ $Ans:~\frac{x^4}{18}+\frac{x^3}{18}+\frac{x^2}{2}+\frac{7 x}{2}+3$