Question

Proving Binomial Theorem by using Mathematical Induction. Step by step.

Answer 1

Hey Again :)
Hmm ok let's state our Binomial Theorem so we know what we're trying to prove.\[\large\rm (x+y)^n\quad=\quad \sum_{k=0}^n \left(\begin{matrix}n \\ \rm k\end{matrix}\right)x^{n-k}y^k\]This thing, ya?

Answer 2

yes

Answer 3

Base Case:\[\large\rm (x+y)^1\quad=\quad x^1+y^1\quad=\quad \sum_{k=0}^1 \left(\begin{matrix}1 \\ \rm k\end{matrix}\right)x^{1-k}y^k\]Does the base case make sense?
Or is how I got from the middle to the right side kinda confusing?

Answer 4

i get it

Answer 5

Induction Hypothesis: We'll assume true for n=m, meaning,\[\large (x+y)^m\begin{align}\quad&=\quad 1x^m+mx^{m-1}y^1+...+mx^{1}y^{m-1}+1y^m\\ &=\sum_{k=0}^m \left(\begin{matrix}m \\k\end{matrix}\right)x^{m-k}y^k\end{align}\]I wrote it both ways, compact and expanded out,
cause I'm not sure which we'll need.

Answer 6

okay

Answer 7

Induction Step: For n=m+1,\[\large\rm (x+y)^{m+1}\quad=\quad \]
Hmm let's see what we can do with this...

Answer 8

\[\large\rm =x^{m+1} ~+~ (m+1)x^{m}y^1 ~+~ ... ~+~(m+1)x^1y^{m} ~+~ y^{m+1}\]So what can we do...
Hmm, thinking...

Answer 9

can we expand (x+y)^m+1 to (x+y)(x+y)^m ?

Answer 10

Ooo we can, yah that's a clever idea

Answer 11

\[\large\rm =\quad (x+y)(x+y)^m\quad=\quad (x+y)\sum_{k=0}^m \left(\begin{matrix}m \\ \rm k\end{matrix}\right)x^{m-k}y^k\]We're allowed to make this step because of our Induction Hypothesis, ya?

Answer 12

yea

Answer 13

Oh oh, and we can also rewrite the first set of brackets using your base case!

Answer 14

ohhhhh i see

Answer 15

\[\large\rm =\quad \color{orangered}{(x+y)^1}(x+y)^m\quad=\quad \color{orangered}{\sum_{k=0}^1\left(\begin{matrix}1 \\ \rm k\end{matrix}\right)x^{1-k}y^k}\sum_{k=0}^m \left(\begin{matrix}m \\ \rm k\end{matrix}\right)x^{m-k}y^k\]

Answer 16

And then um... hmm

Answer 17

hmmm... do we need to use the pascal identity for this question?

Answer 18

Oh the thing for manipulating the coefficients?
Maybe.

Answer 19

maybe we can change (x+y)^m to binomial form

Answer 20

Boy this one is tricky :d hmm

Answer 21

Hmm I found the problem explained on this website:
http://www.math.utah.edu/~bertram/HighSchool/3Induction.pdf
last problem, page 5 and 6

Answer 22

Looks like they just expanded everything out, and then combined like-terms.

Answer 23

I see. I don't like this question :P

Answer 24

i get it tho

Answer 25

Ya no bueno :c