How do you solve #\sum _ { 2} ^ { \infty } 24( 0.8) ^ { k - 1}#?

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Answer 1 · 2017-12-10T02:34:00

#Sum = 96#

Let #Sum = sum_(k=2)^(oo) 24(0.8)^(k-1)#

Apply linerarity.

#Sum = 24*sum_(k=2)^(oo)(0.8)^(k-1)#

Now, let's expand the first few terms of the inifinite sum:

#(Sum)/24 = 0.8^1 + 0.8^2 +0.8^3 + .....#

Here we can see that the RHS is a geometric progression (GP) with first term #(a) = 0.8# and common ratio #(r) = 0.8#

Since #absr <1# the series will converge.

Recall that the sum of a convergent GP is given by: #a/(1-r)#

Hence, in our example,

#(Sum)/24 = 0.8/(1-0.8)#

#Sum = 24 xx 0.8/0.2 = 24 xx 4#

#Sum = 96#