Question 1053828
<pre><b>
Observe that from halfway between 1 and 2, which is {{{1&1/2}}}, 
we go down from {{{1&1/2}}} by {{{1/2}}} to 1, then
we go up from {{{1&1/2}}} by {{{1/2}}} to 2, then
we go down from {{{1&1/2}}} by {{{1/2}}} to 1, then
we go up from {{{1&1/2}}} by {{{1/2}}} to 2, then
etc., etc., etc., ...

We know that the formula {{{(-1)^n}}} goes down and up
-1,1,-1,1,-1,1,...

So we can use that as a multiplier of {{{1/2}}} to go down
and up from {{{1&1/2}}},

So to get the 1st term we take {{{1&1/2}}} and add
{{{(-1)^1}}}, which is -1, times {{{1/2}}}, to go down
to 1.

Then to get the 2nd term we take {{{1&1/2}}} and add
{{{(-1)^2}}}, which is +1, times {{{1/2}}}, to go up
to 2.

Then to get the 3rd term we take {{{1&1/2}}} and add
{{{(-1)^3}}}, which is -1, times {{{1/2}}}, to go down
to 1.

Then to get the 4th term we take {{{1&1/2}}} and add
{{{(-1)^4}}}, which is +1, times {{{1/2}}}, to go up
to 2.

etc., etc., etc.,...

So when n is odd, we go down by {{{1/2}}} to 1 from {{{1&1/2}}}
and when n is even, we go up by {{{1/2}}} to 2 from {{{1&1/2}}}. 

So to get the nth term we take {{{1&1/2}}} and add
{{{(-1)^n}}} times {{{1/2}}},

So the formula for the nth term, which we call {{{a[n]}}} is

{{{matrix(1,7,a[n],"",""="","",1&1/2,""+"",(-1)^n*expr(1/2) )}}}

That will do for the formula, but maybe we want to write
it a little differently. We change {{{1&1/2}}} to an improper
fraction {{{3/2}}}

{{{matrix(1,7,a[n],"",""="","",3/2,""+"",(-1)^n*expr(1/2) )}}}

Then we can factor out {{{1/2}}}

{{{matrix(1,5,a[n],"",""="","",expr(1/2)(3+(-1)^n) )}}}

Then we can change 'multiplying by {{{1/2}}}' to 
'dividing by 2' which is the same thing:

{{{matrix(1,5,a[n],"",""="","",(3+(-1)^n)/2 )}}}

Edwin</pre></b>