Question

help please :) - BINOMIAL THEOREM CC's question below:

Answer 1

Prove:
\[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)3^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)3^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)3^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)3^{8} + 3^{10} = 2^9 (2^{10} +1)\]

Answer 2

I'm considering:
\[(1+x)^{10} = \left(\begin{matrix}10 \\ 0\end{matrix}\right) +\left(\begin{matrix}10 \\ 1\end{matrix}\right)x + \left(\begin{matrix}10 \\ 2\end{matrix}\right)x^2 + \left(\begin{matrix}10 \\ 3\end{matrix}\right)x^3 + ... + x^{10} \]
But i have no idea where to go from here.

If i look at the RHS, then i came up with something....?: \[2^9 ( 2^{10} + 1) \rightarrow 2^{n-1} (2^{n} +1) \]
But i'm not how to get there :/
I also noticed on the LHS, the things are all even. Idk -.-

Answer 3

```
\left(\begin{matrix}10 \\ 2\end{matrix}\right)
```
is better writen as
```
\binom{10}{2}
```
\[\binom{10}{2}\]

Answer 4

oh, thanks for the tip LOL

Answer 5

(x-1)^(10) = cx^10 -cx^9 +cx^8 -cx^7..

right?

Answer 6

what is: (x+1)^(10) + (x-1)^(10)

Answer 7

\[2 (x^{10}+45 x^8+210 x^6+210 x^4+45 x^2+1)\]

Answer 8

ohhh true true . That makes sense!
So that's how they got the even parts only.

Answer 9

im an idiot, so i just use whatever comes to mind :)

Answer 10

oh im not sure how THEY did it .. but this might be a start

Answer 11

yeess.
LOL i can't believe i expanded wrong xD oh gosh.
Okay sooo, what should we do next?

Answer 12

dunno offhand, i was just trying to get your thought to get rid of the odd parts

Answer 13

since LHS and RHS are finite numbers, cant we just let x=3 and work the math to prove?

Answer 14

(x+1)^(10) + (x-1)^(10)

4^(10) + 2^(10)

2^(10) 2^10 + 2^(10)

2^(10) [2^(10)+1]

Answer 15

and divide by 2, since we are twice as heavy

Answer 16

or something like that

Answer 17

.... oh... o.o

Answer 18

can we solve the question like that though? LOL.
I've been trying to sub all kinds of weird numbers everywhere and trying to integrate things -.- I've been getting nowhere tbh LOl

Answer 19

direct solutions tend to work out great to me ...

\[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)x^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)x^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)x^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)x^{8} + x^{10} =\frac{(x+1)^{10}+(x-1)^{10}}{2}\]

Answer 20

when x=3, ive already worked the numerator to 2^(10) [2^(10)+1]

right?

Answer 21

yes

Answer 22

then it is proofed

Answer 23

hmm okay then. Thank you!
It's just that for the prev. questions i've been working on, i've had to use a range of differentiation/integration and substitution methods so i thought this question would've involved similar steps but thanks again!
Your method was way simpler.

Answer 24

\[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)3^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)3^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)3^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)3^{8} + 3^{10}\]
\[\frac{(3+1)^{10}+(3-1)^{10}}{2}\]
\[\frac{4^{10}+2^{10}}{2}\]
\[\frac{2^{10}(2^{10}+1)}{2}\]
\[{2^{9}(2^{10}+1)}\]

Answer 25

well, it was your way, i just built on it :)

Answer 26

aahh okay! Thanks again :D!