## TrojanPoem one year ago Prove that: 2 * nC0 + 2^2 * nC1 / 2 + 2^3 * nC2/3 + 2^4 * nC3/4 + ... + 2^(n+1) * nCn / n+1 = 3^(n+1) -1 / n+1

1. ganeshie8

use latex friend

2. TrojanPoem

@dan815 , Hmm ganshie ?

3. TrojanPoem

What is latex friend ?

4. dan815

where you type \\ [ \] before and after

5. dan815

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6. dan815

wow thats quite a find

7. ganeshie8

expand $$\large \dfrac{(1+2)^n}{n}$$ using binomial thm ?

8. dan815

oh ya xD

9. TrojanPoem

$\frac{ 2 C_{0}^{n} }{ 1 } + \frac{ 2^2 C_{1}^{n} }{ 2 } + \frac{ 2^3 C_{3}^{n} }{ 3 } + \frac{ 2^4 C_{4}^{n} }{ 4} + ... + \frac{ 2^{n+1} C_{n}^{n} }{ n+1 } = \frac{ 3^{n+1} -1 }{ n+1 }$

10. dan815

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11. dan815

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12. dan815

they seem so close

13. dan815

for startes thats just a constant we can put into the equation so

14. dan815

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15. TrojanPoem

Do you really have to use this E ? I hate it.

16. dan815

it just means u increment k by 1 everytime and add the new expression, id have to copy paste so much if we dont use it

17. dan815

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18. dan815

thats all i means

19. dan815

here is somethign useful

20. dan815

i think i saw this one identity before equation a summation of the summation of choose for 1 less with respect to +1 choosing

21. dan815

let me find it, its in an older question i asked

22. TrojanPoem

Ok , brb Until you find it.

23. dan815

ok here are some useful ones we can go over the proofs for them depending on which ones we use

24. TrojanPoem

Ok

25. dan815

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26. dan815

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27. dan815

the 2nd one is the one we just worked out fromm ganeshies suggesntion

28. dan815

cause you think about expanding this bracket 2^n = (1+1)^n, this will just be that summation right

29. dan815

oh baby i got an idea!! we can do those geometric series tricks here!! multiply by k and subtract the 2 sequences it will simplify i think

30. TrojanPoem

show me

31. dan815

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32. dan815

can u show that one first, that is almost the same

33. dan815

@ganeshie8

34. TrojanPoem

I think ganshie is busy

35. dan815

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36. dan815

ehh im chucking this xD

37. dan815

wait not yet it might bw a useful idea

38. dan815

if we can break it down into summations we know like some cobinations of the 2^n summation it looks so close to the gemetric series tricks

39. ganeshie8

how about using derivatives, the right hand side is the derivative of x^n when x=2

40. ganeshie8

$\large (1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$ differentiate both sides and get $\large n(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k}kx^{k-1}$ let $$x=1$$