How do you use summation notation to expression the sum #32+24+18+...+10.125#?

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Answer 1 · 2017-03-09T12:54:48

#sum_(n = 0)^4 32cdot (3/4)^n#

#32+24+18+...+10.125#

The initial pattern is:

#32* (3/4)^0+24* (3/4)^1+18* (3/4)^2+...+10.125#

And therefore:

#10.125 = 32* (3/4)^n#

#implies n = ln(10.125/32)/ln(3/4) = n = 4#

So it's:

#32* (3/4)^0+32* (3/4)^1+32* (3/4)^2+...+32* (3/4)^4#

We can say then that the series sum is:

#sum_(n = 0)^4 32cdot (3/4)^n#