How to show that #S_oo- S_n= (-1/3)^n# ,given, #S_oo=9/4# and #S_n=9/4(1-(-1/3)^n)#?

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Answer 1 · 2018-06-21T08:35:29

I think there's a #9/4# factor missing

Since you're given both #S_\infty# and #S_n#, in order to compute #S_\infty - S_n# you can simply subtract the two expressions:

#color(red)(S_\infty)-color(blue)(S_n) = color(red)(9/4) - color(blue)(9/4(1-(-1/3)^n)#

We can expand #S_n# as follows:

#9/4(1-(-1/3)^n) = 9/4 - 9/4\times(-1/3)^n#

So, we have

#S_\infty-S_n#
# = 9/4-(9/4 - 9/4\times(-1/3)^n)#
# = cancel(9/4)-cancel(9/4) + 9/4\times(-1/3)^n#

So,

#S_\infty-S_n =9/4\times(-1/3)^n#