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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
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

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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 , Hmm ganshie ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0What is latex friend ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2where you type \\ [ \] before and after

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0expand \(\large \dfrac{(1+2)^n}{n}\) using binomial thm ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2 C_{0}^{n} }{ 1 } + \frac{ 2^2 C_{1}^{n} }{ 2 } + \frac{ 2^3 C_{3}^{n} }{ 3 } + \frac{ 2^4 C_{4}^{n} }{ 4} + ... + \frac{ 2^{n+1} C_{n}^{n} }{ n+1 } = \frac{ 3^{n+1} 1 }{ n+1 } \]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for startes thats just a constant we can put into the equation so

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Do you really have to use this E ? I hate it.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.2here is somethign useful

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i think i saw this one identity before equation a summation of the summation of choose for 1 less with respect to +1 choosing

dan815
 one year ago
Best ResponseYou've already chosen the best response.2let me find it, its in an older question i asked

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Ok , brb Until you find it.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ok here are some useful ones we can go over the proofs for them depending on which ones we use

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the 2nd one is the one we just worked out fromm ganeshies suggesntion

dan815
 one year ago
Best ResponseYou've already chosen the best response.2cause you think about expanding this bracket 2^n = (1+1)^n, this will just be that summation right

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2can u show that one first, that is almost the same

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I think ganshie is busy

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ehh im chucking this xD

dan815
 one year ago
Best ResponseYou've already chosen the best response.2wait not yet it might bw a useful idea

dan815
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.0how about using derivatives, the right hand side is the derivative of x^n when x=2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\large (1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k\] differentiate both sides and get \[\large n(1+x)^{n1}=\sum_{k=0}^n \binom{n}{k}kx^{k1}\] let \(x=1\)
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