How can I find the Sum of #5-10/3+20/9-40/27+80/81-#... ?

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Answer 1 · 2017-11-15T22:33:05

#5-10/3+20/9-40/27+80/81-... = 3#

Assuming the general term is:

#a_n = (-1)^n5*2^n/3^n#

we have:

#sum_(n=0)^oo (-1)^n5*2^n/3^n = 5 sum_(n=0)^oo (-2/3)^n#

this is the sum of a geometric series of ratio #q=-2/3# and we now that for #absq < 1#

#sum_(n=0)^oo q^n = 1/(1-q)#

so:

#sum_(n=0)^oo (-1)^n5*2^n/3^n = 5/(1+2/3) = 3#