I have tried a wide variety of exponential notation, logirithms, etc.  Any help to get moving in the right direction would be helpful
asked May 3, 2016 in Algebra 1 Answers by anonymous

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Part of the answer is ∑1/(3^k+√(k+1)) for k=1 to n. If we put k=0, we get 1/(1+1)=1/2 or 0.5. This does not fit in with the first terms. It seems only to apply when k>0.

Take the general term of the series. The fraction can be rationalised: 1/(3^k+√(k+1) * (3^k-√(k+1))/(3^k-√(k+1)). This rationalises to (3^k-√(k+1))/(3^2k-k-1). For k=3 this is  (27-2)/(729-4)=25/725=1/29=1/(27+2). For k=1, rationalisation produces (3-√2)/(9-2)=(3-√2)/7. Note that the denominator is always rational. The first three terms 4+1/5+0.3=4.5. As it happens this can be split into 4 +1/2, where 1/2 is the original summation starting with k=0. The series then becomes:

4+∑1/(3^k+√(k+1)) for k=0 to n; or 4+∑(3^k-√(k+1))/(3^2k-k-1) in rational form (use other form when k=0 to avoid division by zero).


answered May 3, 2016 by Rod Top Rated User (582,800 points)

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