How do you find #S_n# for the geometric series #a_1=4#, r=-3, n=5?

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Answer 1 · 2016-10-18T17:16:33

#S_5=244#

Here in given geometric series #a_1=4# and #r=-3#

Now if #a_1=a#, #S-n# the sum of geometric series is

#S_n=a+ar+ar^2+.................+ar^(n-1)# and therefore

#rS_n=ar+ar^2+.................+ar^(n-1)+ar^n#

and subtracting first from second #S_n(r-1)=a(r^n-1)#

and #S_n=(a(r^n-1))/(r-1)#

Hence in the given geometric series

#S_5=(4((-3)^5-1))/(-3-1)#

= #(4(-243-1))/(-4)#

=#244#