Serie and sequencesss (1 Viewer)

[angeelaa]

Hey guyss
how do you solve 2.81818181.... as a fraction using the limiting sum method??

 

[anomalousdecay]

Let x = 2.818181...

Multiply x by 100.

You get 281.81818...

Now take away 2.8181818.... = x from 100x

You get 99x = 279

Therefore we get x = 279/99 (simplify it in calculator or whatever).

If you have one digit recurring, then you multiply by 10, 2 digits recurring multiply by 100, 3 digits recurring you multiply by 1000 = 10^3, etc.

 

[angeelaa]

omgg thank you! you're a lifesaver

 

[Carrotsticks]

Let x = 2.818181...

Multiply x by 100.

You get 281.81818...

Now take away 2.8181818.... = x from 100x

You get 99x = 279

Therefore we get x = 279/99 (simplify it in calculator or whatever).

If you have one digit recurring, then you multiply by 10, 2 digits recurring multiply by 100, 3 digits recurring you multiply by 1000 = 10^3, etc.

This is not using the limiting sum method, however. A student using this method in a "Series and Sequences" test will most certainly get penalised.

 

[faisalabdul16]

I thought the way to do this question using series and sequences was to split the number into 2 and 0.8181818181 and then
0.81818181
is the same as 81/100 + 81/10000 + 81/1000000 + .... this ends up being a GP, and then you use the limiting sum formula , a/1-r
which gives 9/11.
Then 2+9/11 = 31/11 = 2.81818181...

 

[Carrotsticks]

I thought the way to do this question using series and sequences was to split the number into 2 and 0.8181818181 and then
0.81818181
is the same as 81/100 + 81/10000 + 81/1000000 + .... this ends up being a GP, and then you use the limiting sum formula , a/1-r
which gives 9/11.
Then 2+9/11 = 31/11 = 2.81818181...

Correct. This is what they look for.