Tuesday, 14 June 2016

calculus - Solving the infinite sum suminftyk=0left(frac13right)k+3



I'm stuck on the question k=0(13)k+3



I know that k=0(13)k is solved by using k=0ak= 11a and the answer is 32



So is there a way I could apply that to the above question or is there a different way to approach the problem?



Answer



nk=0(13)k+3=133nk=0(13)k1271113=12732=118


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