TrojanPoem
  • TrojanPoem
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
Mathematics
katieb
  • katieb
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ganeshie8
  • ganeshie8
use latex friend
TrojanPoem
  • TrojanPoem
@dan815 , Hmm ganshie ?
TrojanPoem
  • TrojanPoem
What is latex friend ?

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dan815
  • dan815
where you type \\ [ \] before and after
dan815
  • dan815
|dw:1435164810600:dw|
dan815
  • dan815
wow thats quite a find
ganeshie8
  • ganeshie8
expand \(\large \dfrac{(1+2)^n}{n}\) using binomial thm ?
dan815
  • dan815
oh ya xD
TrojanPoem
  • TrojanPoem
\[\frac{ 2 C_{0}^{n} }{ 1 } + \frac{ 2^2 C_{1}^{n} }{ 2 } + \frac{ 2^3 C_{3}^{n} }{ 3 } + \frac{ 2^4 C_{4}^{n} }{ 4} + ... + \frac{ 2^{n+1} C_{n}^{n} }{ n+1 } = \frac{ 3^{n+1} -1 }{ n+1 } \]
dan815
  • dan815
|dw:1435164975210:dw|
dan815
  • dan815
|dw:1435165082146:dw|
dan815
  • dan815
they seem so close
dan815
  • dan815
for startes thats just a constant we can put into the equation so
dan815
  • dan815
|dw:1435165254017:dw|
TrojanPoem
  • TrojanPoem
Do you really have to use this E ? I hate it.
dan815
  • 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
dan815
  • dan815
|dw:1435165420765:dw|
dan815
  • dan815
thats all i means
dan815
  • dan815
here is somethign useful
dan815
  • 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
dan815
  • dan815
let me find it, its in an older question i asked
TrojanPoem
  • TrojanPoem
Ok , brb Until you find it.
dan815
  • dan815
ok here are some useful ones we can go over the proofs for them depending on which ones we use
TrojanPoem
  • TrojanPoem
Ok
dan815
  • dan815
|dw:1435165847184:dw|
dan815
  • dan815
|dw:1435165991834:dw|
dan815
  • dan815
the 2nd one is the one we just worked out fromm ganeshies suggesntion
dan815
  • dan815
cause you think about expanding this bracket 2^n = (1+1)^n, this will just be that summation right
dan815
  • 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
TrojanPoem
  • TrojanPoem
show me
dan815
  • dan815
|dw:1435166233288:dw|
dan815
  • dan815
can u show that one first, that is almost the same
dan815
  • dan815
TrojanPoem
  • TrojanPoem
I think ganshie is busy
dan815
  • dan815
|dw:1435166298788:dw|
dan815
  • dan815
ehh im chucking this xD
dan815
  • dan815
wait not yet it might bw a useful idea
dan815
  • 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
ganeshie8
  • ganeshie8
how about using derivatives, the right hand side is the derivative of x^n when x=2
ganeshie8
  • 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 \(x=1\)

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