as with any geometric sequence,
Sn = a(1-r^n)/(1-r)
So, since a=x and r=x,
Sn = x(1-x^n)/(1-x)
1b) S = a/(1-r) = x/(1-x)
1c should not be too hard now.
an exponential sequence (geometric progression) has first term x and common ratio x ,x≠1 write down the sum of the first n terms of this sequence
(1a): write down the sum of the first n terms of this sequence
(1b): |x|<1 find lim Sn as n=>infinity
(1c): if An=sum r=1 to n Sr,show that
(1-x)An=nx-xSn
Please help me show working this one difficult for me
1 answer
1 answer