Why is Sequence 1, 4, 7, 10 Written as 3n-2 or 3n+2?

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Hi,

Could anyone please explain to me very simply why in a sequence say 1, 4, 7, 10,... which has the general term form: a_n = 3n-2 can also be written as 3n+2?

Why do some people use 3n+2 rather than 3n-2, what advantage has that got? Is it actually more mathematically correct to write 3n+2?

So 2n-2 or 2n+2 would be for a sequence 1, 2, 3, 4, 5,...
4n-1 or 4n+3 would be for 3, 7, 11, ...

Thanks

Nat
 
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Surely you mean 3n+1 with n starting from 0??
 
If an= 3n-2 then 3(n+1)- 2= 3n+3-2= 3n+1. The only difference is that with an= 3n-2 you have a1= 1, a2= 4, etc. while with an= 3n+1 it is a0= 1, a1= 4, etc. Just a difference in where you start counting.
There is no real advantage- just that some people don't like to start counting with 0!


"So 2n-2 or 2n+2 would be for a sequence 1, 2, 3, 4, 5,... "

No- 2(1)-2= 0, 2(2)-2= 2, but 2(3)- 2= 4 not 3. It should be obvious that 2n- 2 and 2n+ 2 are always even numbers. Did you mean 2, 4, 6, ...?


"4n-1 or 4n+3 would be for 3, 7, 11, ..."
Yes, one starts with n= 1, the other with n= 0.
 
HallsofIvy said:
If an= 3n-2 then 3(n+1)- 2= 3n+3-2= 3n+1. The only difference is that with an= 3n-2 you have a1= 1, a2= 4, etc. while with an= 3n+1 it is a0= 1, a1= 4, etc. Just a difference in where you start counting.
There is no real advantage- just that some people don't like to start counting with 0!


"So 2n-2 or 2n+2 would be for a sequence 1, 2, 3, 4, 5,... "

No- 2(1)-2= 0, 2(2)-2= 2, but 2(3)- 2= 4 not 3. It should be obvious that 2n- 2 and 2n+ 2 are always even numbers. Did you mean 2, 4, 6, ...?


"4n-1 or 4n+3 would be for 3, 7, 11, ..."
Yes, one starts with n= 1, the other with n= 0.

Yes sorry I did mean 2, 4, 6, 8, ...
 
I don't know about HallsofIvy, but I'm an inveterate "from zero"-counter..
 
Last edited:
arildno said:
I don't know about HallsofIvy, but I'm an inveterate "from zero"-counter..

Unfortunately (or fortunately depending on your preference), many textbooks count from 1.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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