Let any position count be #i#
Let any term be #a_i#
So
#a_1->" first term"#
#a_2->" second term"#
#a_i->" "ith" term"#
Two things I notice:
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Point 1:")->#Alternates between positive and negative starting at positive.
So this can be achieved by
#a_i->a_1 xx(-1)^2#
#a_i->a_2xx(-1)^3#
#a_i->a_3xx(-1)^4#
#=>a_i->a_ixx(-1)^(i+1)#
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Point 2:")->#Each term is half the previous
So we have
#(-1)^2xxa_1#
#a_2=(-1)^3xx1/2xxa_1#
#a_3=(-1)^4xx1/2xx1/2xxa_1#
This implies #a_i=(-1)^(i+1)xx(1/2)^(i-1)xxa_1#
Where #a_1=+896#
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)(a_i=896(-1)^(i+1)(1/2)^(i-1)#
Thus
#" "color(purple)(bar(ul(|color(white)(2/2)a_8=896(-1)^(8+1)(1/2)^(8-1) = -7color(white)(2/2)|)))#