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Find the sum of 1/6*15 + 1/15*24 +1/24*33 + 1/33*42 + ... I have no idea how to relate this series mathematically and am driving myself crazy trying to figure it out. I know that each of the multiplied terms has a difference of 9, and that the first term in each successive fraction is the second term of the previous one, but I cannot figure out how to make that information useful. Any ideas?
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\[\sum_{n=1}^{\infty} (1/((9n-3)*(9n+6))\] it may be like this
Thank you so much. It was driving me bonkers trying to figure it out!
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