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Rufus has the right idea.
((1/3)^t)*cos(wt) is the real part of [(1/3)e^(iw)]^t. So the sum is the real part of (1/3)e^(iw)/[1-(1/3)e^(iw)], using the formula for the sum of a geometric series. Now substitute e^(iw)=cos w + i sin w, and use the fact that the real part of (a+ib)/(c+id) is (ac+bd)/(c^2+d^2). When I do the algebra, I get (1-3cos(w))/(3cos(w)-5), which is what you wanted to show. |
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