Question 626875
1.1 To get the next term, multiply by {{{(1/2)*(x-2)}}}. This is the common ratio, r. The first term is {{{a[1] = 16(x-2)^3}}}. The sum of the infinite geometric series is equal to
 {{{S = a[1]/(1-r)}}}
 
{{{16(x-2)^3+8(x-2)^4+4(x-2)^5+...}}} = {{{16(x-2)^3/(1-(1/2)*(x-2)) }}}
                                      = {{{32(x-2)^3/(4-x)) }}}
1.1.1 The geometric series is convergent iff {{{abs(r) < 1}}}
{{{abs((1/2)*(x-2))<1}}}
{{{ -1 < (1/2)*(x-2) < 1}}}
{{{-2 < x-2 < 2}}}
{{{highlight(0<x<4)}}} 

1.1.2 x=2.5
{{{32(x-2)^3/(4-x)) }}}, substitute x = 2.5
 = {{{32(2.5-2)^3/(4-2.5)) }}}
={{{highlight(8/3)}}}