Question 696784
Let the numbering of the terms be
1,2,3,4, . . . etc.
The changing sign part goes
term #1 is (-)
term #2 is (+)
term #3 is (-)
etc.
That part of the general formula is
{{{ (-1)^n }}} where {{{ n }}} is the number of the term
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It looks like I might be able to have a {{{ 1/2 }}} in each
term, so i've got so far:
{{{ (1/2)*(-1)^n }}}
But, obviously, that's not all of it.
What if I put in a factor of {{{ 3^( 3-n) }}} ? Now I have
{{{ (1/2)*( 3^(3-n) )*(-1)^n }}}
I'll test it out
{{{ n = 1 }}}
{{{ (1/2)*( 3^(3-1))*(-1)1^1 = (1/2)*9*(-1) }}}
{{{ -9/2 }}}
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{{{ n = 2 }}}
{{{ (1/2)*( 3^(3-2))*(-1)1^2 = (1/2)*3*1 }}}
{{{ 3/2 }}}
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{{{ n = 3 }}}
{{{ (1/2)*( 3^(3-3))*(-1)1^3 = (1/2)*3^0*(-1) }}}
{{{ -1/2 }}}
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{{{ n = 4 }}}
{{{ (1/2)*( 3^(3-4))*(-1)1^4 = (1/2)*3^(-1)*1 }}}
{{{ 1/6 }}}
This works
So the formula is
{{{ (1/2)*( 3^(3-n) )*(-1)^n }}}
I'm  not sure about the summation notation. Are they 
looking for the sum of all the terms?
Us the sigma ( summation ) symbol in front
of what I just gave you.