That would take a long time for us to calculate!!Instead,let's try something else.

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Let \(A=\dfrac{2}{1*3}+\dfrac{2}{3*5}+\dfrac{2}{5*7}+\dfrac{2}{7*9}+\dfrac{2}{9*11}\) and let \(B=\dfrac{1}{1*2}+\dfrac{1}{2*3}+\dfrac{1}{3*4}+\dfrac{1}{4*5}+\dfrac{1}{5*6}+\dfrac{1}{6*7}+\dfrac{1}{7*8}\),then evaluate \(33A-32B\).

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