How do you find the sum of the finite geometric sequence of #sum_(j=1)^12 16(1/2)^(j-1)#?

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Answer 1 · 2017-07-13T17:35:28

Given: #sum_(j=1)^12 16(1/2)^(j-1)#

This can be written as:

#16+ sum_(j=1)^11 16(1/2)^j" [1]"#

From this reference we obtain the formula:

#S_n = sum_(j=1)^na_j= a_1(1-r^n)/(1-r)#

The formula fits the second term of equation [1] where #a_1 = 16, r=1/2 and n = 11#

#S_12 = 16+ 16(1-(1/2)^11)/(1-(1/2))#

#S_12 = 47.984375#