Prove the divisibility

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Randyyy

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show that (2n+1)3n-1 is NOT divisible by 4 for every integer [MATH]n \geq 0[/MATH] ...................................edited
2n*3n-1+3n-1-1 [MATH]\equiv[/MATH]2n(-1)-1n-1-1
2n(-1)-1n-1-1 [MATH]\equiv[/MATH] -2n-2 [MATH]\equiv[/MATH] 2n+2 [MATH]\equiv[/MATH] 2(n+1)
But this is not true as it won´t be divisible by 4 for values of n=1 and n=2.
 
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"show that \(\displaystyle (2n+1)3^{n-1}\) is divisible by 4 for every integer n≥0".
Yes, that statement is simply not true for n= 0, 1, 2, 3, etc.
Are you sure the problem was not "show that \(\displaystyle (2n+1)3^{n-1}\) is NOT divisible by 4 for every integer n≥0? both 2n+1 and \(\displaystyle 3^{n-1}\) are odd for all n so their product is not divisible by 2 so certainly not divisible by 4..
 
you´re right, I apologize, it was suppoed to be: show that (2n+1)3n-1 is divisible by 4 for every interger [MATH] n \geq 0[/MATH](2n+1)3n-1 [MATH]\equiv[/MATH] 2n*3n+3n-1
2n*3n+3n-1 [MATH]\equiv[/MATH] 2n(-1n)-1n-1 [MATH]\equiv[/MATH] -2n-2
-2n-2 [MATH]\equiv[/MATH] 2n+2 [MATH]\equiv[/MATH] 2(n+1)
 
Randyyy said:
you´re right, I apologize, it was suppoed to be: show that (2n+1)3n-1 is divisible by 4 for every interger [MATH] n \geq 0[/MATH](2n+1)3n-1 [MATH]\equiv[/MATH] 2n*3n+3n-1
2n*3n+3n-1 [MATH]\equiv[/MATH] 2n(-1n)-1n-1 [MATH]\equiv[/MATH] -2n-2
-2n-2 [MATH]\equiv[/MATH] 2n+2 [MATH]\equiv[/MATH] 2(n+1)
Click to expand...
Is it:

\(\displaystyle (2n+1) * 3^{n} -1\) ....................................this is what you wrote in response #3

or

\(\displaystyle (2n+1) * 3^{n-1} \) ....................................this is what Prof. Ivey suggested in response #2 and you wrote in response #1

Which expression do you have to manipulate?
 
Subhotosh Khan said:
Is it:

\(\displaystyle (2n+1) * 3^{n} -1\) ....................................this is what you wrote in response #3

or

\(\displaystyle (2n+1) * 3^{n-1} \) ....................................this is what Prof. Ivey suggested in response #2 and you wrote in response #1

Which expression do you have to manipulate?
Click to expand...
It was supposed to be \(\displaystyle (2n+1) * 3^{n}-1 \). I accidentally put the -1 in the exponent.
 
Randyyy said:
show that (2n+1)3n-1 is NOT divisible by 4 for every integer [MATH]n \geq 0[/MATH] ...................................edited
2n*3n-1+3n-1-1 [MATH]\equiv[/MATH]2n(-1)-1n-1-1
2n(-1)-1n-1-1 [MATH]\equiv[/MATH] -2n-2 [MATH]\equiv[/MATH] 2n+2 [MATH]\equiv[/MATH] 2(n+1)
But this is not true as it won´t be divisible by 4 for values of n=1 and n=2.
Click to expand...
2n*3n-1+3n-1-1 ≡ 2n(-1)-1n-1-1 ....... How did you calculate this equivalence?

2n(-1)-1n-1-1 ≡ -2n-2 [MATH]\equiv[/MATH] 2n+2 [MATH]\equiv[/MATH] 2(n+1)....... How did you calculate this equivalence?

But this is not true as it won´t be divisible by 4 for values of n=1 and n=2.
 
I see my mistake, it is quite a stupid one, I did it on both the highlighted expressions.
I thought that (-1)n=-1, but this is clearly not true because if n is even then (-1)n=1

So it should be: (2n+1)*3n- 1[MATH]\equiv[/MATH] 2n(-1)n+(-1)n-1.But I am unsure what I should go from there. should I make n=2k, n=k and check if it is true for odd and even exponents?
 
You wrote -1n = -1 for all n. Why? 1n=1 for all n so -1n=-1 for all n.
Now (-1)n is completely different. It is 1 for n even and -1 for n odd.
 
Randyyy said:
I see my mistake, it is quite a stupid one, I did it on both the highlighted expressions.
I thought that (-1)n=-1, but this is clearly not true because if n is even then (-1)n=1

So it should be: (2n+1)*3n- 1[MATH]\equiv[/MATH] 2n(-1)n+(-1)n-1.But I am unsure what I should go from there. should I make n=2k, n=k and check if it is true for odd and even exponents?
Click to expand...
So please put up corrected "work" with detailed steps described.
 
Randyyy said:
It was supposed to be \(\displaystyle (2n+1) * 3^{n}-1 \). I accidentally put the -1 in the exponent.
Click to expand...


\(\displaystyle (2n + 1)*3^n \ - \ 1 \ \) IS divisible by 4 for every integer \(\displaystyle \ n \ge 0. \)

I verified it for myself with a proof. It would make sense that this is the problem,
as this is not relatively trivial.
 
After correcting my previous error I conlcluded this;
(2n+1)*3n-1=2n*3n+3n-1
2n*3n+3n-1[MATH]\equiv[/MATH] 2n(-1)n+(-1)n-1 (mod 4)
Case 1: n is odd:
2(2k+1)(-1)-2=-4k-4 [MATH]\equiv 0[/MATH] (mod 4)
Case 2: n is even:
4k+1-1[MATH]\equiv 0[/MATH] (mod4)

I also wanna add that I wanna excuse the confusion I caused with my previous posts, thank for all the input!
 
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