With "n" being a positive integer, evaluate the sum :
C(n,0)+2C(n,1)+2^2C(n,2)+...+2kC(n,k)+...+2nC(n,n)
In the book it gives me the answer (1 + 2)^n = 3^n, which I know is the binomial expansion of that sum, however I was not able arrive at that answer
. In short, I know that we're supposed to use the Binomial Theorem and plug in values, but I'm not sure what I have to plug in to X AND Y to get the answer Also please look at this.
[FONT=MathJax_Size1][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Size1]∑[/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Math](n[/FONT][FONT=MathJax_Math]k)[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]^k[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]^n[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]^n[/FONT]I managed to find these steps to my question on a forum, but there was no explanation to how the guy managed to figure out x = 2 and y = 1. I really need to know how it was figured out that x =2 and y=1. Please don't tell me that the only way to get Y and X is to guess, I'm sure there was a logical way he figured out the numbers to plug in.
[FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Size1][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT]
C(n,0)+2C(n,1)+2^2C(n,2)+...+2kC(n,k)+...+2nC(n,n)
In the book it gives me the answer (1 + 2)^n = 3^n, which I know is the binomial expansion of that sum, however I was not able arrive at that answer
. In short, I know that we're supposed to use the Binomial Theorem and plug in values, but I'm not sure what I have to plug in to X AND Y to get the answer Also please look at this.[FONT=MathJax_Size1][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Size1]∑[/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Math](n[/FONT][FONT=MathJax_Math]k)[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]^k[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]^n[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]^n[/FONT]I managed to find these steps to my question on a forum, but there was no explanation to how the guy managed to figure out x = 2 and y = 1. I really need to know how it was figured out that x =2 and y=1. Please don't tell me that the only way to get Y and X is to guess, I'm sure there was a logical way he figured out the numbers to plug in.
[FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Size1][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT]
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