SOLUTION: Find the sum of the infinite series: ∞ ∑ (3/2^i+4/5^i) i=1 So far I know that it goes: 23/10, 9

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Question 204345: Find the sum of the infinite series:
∞
∑ (3/2^i+4/5^i)
i=1
So far I know that it goes: 23/10, 91/100, 407/1000, etc., in multiples of ten for the denominator, but I am confused as to what the top sequence is and if it's supposed to go to infinity, and then how would I find the sum?
Thank you in advance
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sum%283%2F%282%5Ei%29%2B4%2F%285%5Ei%29%2Ci=1%2Cinfinity%29 Start with the given summation


sum%283%281%2F%282%5Ei%29%29%2B4%281%2F%285%5Ei%29%29%2Ci=1%2Cinfinity%29 Break up the fractions


Rewrite 1 as 1%5Ei. Note: 1%5Ei=1 for all values of 'i'


sum%283%281%2F2%29%5Ei%2B4%281%2F5%29%5Ei%2Ci=1%2Cinfinity%29 Use the identity %28x%5Ez%29%2F%28y%5Ez%29=%28x%2Fy%29%5Ez


sum%283%281%2F2%29%5Ei%2Ci=1%2Cinfinity%29%2Bsum%284%281%2F5%29%5Ei%2Ci=1%2Cinfinity%29 Break up the summation.


Re-index the summations to start them at zero. Note: this will replace each 'i' with 'i+1' since this offset occurs.


Pull out the first terms of each summation. We're subtracting them off since we're originally starting at i=1 anyway.


Raise each term to the 0th power to get 1.


sum%283%281%2F2%29%5Ei%2Ci=0%2Cinfinity%29-3%2Bsum%284%281%2F5%29%5Ei%2Ci=0%2Cinfinity%29-4 Reduce


Now recall that the sum "S" for an infinite geometric series a%5Bn%5D=a%2Ar%5En is S=a%2F%281-r%29. So this means that...


3%2F%281-1%2F2%29-3%2B4%2F%281-1%2F5%29-4 Replace the summations with the given sum formulas (see above)


3%2F%281%2F2%29-3%2B4%2F%284%2F5%29-4 Subtract


3%2A%282%2F1%29-3%2B4%2A%285%2F4%29-4 Multiply by the reciprocal


6%2F1-3%2B20%2F4-4 Multiply


6-3%2B5-4 Reduce


4 Combine like terms.


So sum%283%2F%282%5Ei%29%2B4%2F%285%5Ei%29%2Ci=1%2Cinfinity%29=4


Note: if you add up the pieces of this infinite summation, you'll find that the sums will get closer and closer to 4 (you might have to find more terms).