Cancel f(2) from both the sides and then compare:
f(1)< f(3)

Given that \(f(n)= \frac{n}{n+1}\) and we need to find the value of f(1) × f(2) and f(2) × f(3)

Since, both the sides have f(2), so we can cancel it out and then compare only f(1) and f(3)

To find f(1) we need to compare what is inside the bracket in f(1) and f(n)

=> We need to substitute n with 1 in \(f(n)= \frac{n}{n+1}\) to get the value of f(1)

=> \(f(1)= \frac{1}{1+1}\) = \(\frac{1}{2}\)

Similarly, \(f(3)= \frac{3}{3+1}\) = \(\frac{3}{4}\)

Quantity A = f(1) = \(\frac{1}{2}\) = 0.5

Quantity B = f(3) = \(\frac{3}{4}\) = 0.75

Clearly, Quantity B > Quantity A

So, Answer will be B
Hope it helps!

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