How do you evaluate #\sum _ { n = 1} ^ { 9} ( - 4) ^ { n - 1}#?

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Answer 1 · 2017-10-04T11:38:21

# sum^9 =52429#
#n=1#

# sum^9 (-4)^(n-1) ; n= 1 ; (-4)^(n-1)=(-4)^0 =1 #
n=1
#n= 2 ; (-4)^(n-1)=(-4)^1 = -4 #

#n= 3 ; (-4)^(n-1)=(-4)^2 = 16#

#n= 4 ; (-4)^(n-1)=(-4)^3 = -64#

So the terms are #{ 1 , -4 ,16 ,-64.....}# . This is geometric

progression series of which first term is #a_1= 1# and common

ratio is # r= 16/-4= -4/1 = -4# and number of terms is #n=9#

# sum^9 = a_1*(r^n-1)/(r-1) = 1 * ((-4)^9-1)/(-4-1) =52429#
#n=1#
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