## butterflydreamer one year ago help please :) - BINOMIAL THEOREM CC's question below:

1. butterflydreamer

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

2. butterflydreamer

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 -.-

3. amistre64

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

4. butterflydreamer

oh, thanks for the tip LOL

5. amistre64

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

6. amistre64

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

7. amistre64

$2 (x^{10}+45 x^8+210 x^6+210 x^4+45 x^2+1)$

8. butterflydreamer

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

9. amistre64

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

10. amistre64

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

11. butterflydreamer

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

12. amistre64

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

13. amistre64

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

14. amistre64

(x+1)^(10) + (x-1)^(10) 4^(10) + 2^(10) 2^(10) 2^10 + 2^(10) 2^(10) [2^(10)+1]

15. amistre64

and divide by 2, since we are twice as heavy

16. amistre64

or something like that

17. butterflydreamer

.... oh... o.o

18. butterflydreamer

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

19. amistre64

direct solutions tend to work out great to me ... $1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)x^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)x^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)x^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)x^{8} + x^{10} =\frac{(x+1)^{10}+(x-1)^{10}}{2}$

20. amistre64

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

21. butterflydreamer

yes

22. amistre64

then it is proofed

23. butterflydreamer

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.

24. amistre64

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

25. amistre64

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

26. butterflydreamer

aahh okay! Thanks again :D!