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If f:N→N, where N is set of natural numbers and f(xy)=f(x)⋅f(y)−f(x+y)+1 ∀x,y∈N and f(1)=2, then the value of 1+∞∑k=1f(k)2k is

Answer» If f:NN, where N is set of natural numbers and f(xy)=f(x)f(y)f(x+y)+1 x,yN and f(1)=2, then the value of 1+k=1f(k)2k is


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