Question

How do you find the nth term of the sequence `0.6, 0.61, 0.616, 0.6161,...`?

Answer 1

`a_n = (122-10^(-n)(77+(-1)^n*45))/198`

Explanation:

Note that:

`0.bar(61) = 61/99`

`0.bar(16) = 16/99`

So we find:

`0.6 = 0.bar(61)-1/10*0.bar(16) = (61-10^(-1) * 16)/99`

`0.61 = 0.bar(61)-1/100*0.bar(61) = (61-10^(-2)*61)/99`

`0.616 = 0.bar(61)-1/1000*0.bar(16) = (61-10^(-3)*16)/99`

`0.6161 = 0.bar(61)-1/10000*0.bar(61) = (61-10^(-4)*61)/99`

. . .

We can match this alternating pattern using `(-1)^n`...

`a_n = (122-10^(-n)(77+(-1)^n*45))/198`