Telangana TET Paper 1 Exam Model Paper 1

Concept:Each term $(1+\frac{1}{n})$ simplifies to $\frac{n+1}{n}$. Multiplying them creates a telescoping product where most numerators and denominators cancel.Explanation:Rewrite every factor: $(1+\frac{1}{3}) = \frac{4}{3}$, $(1+\frac{1}{4}) = \frac{5}{4}$, $\ldots$, $(1+\frac{1}{98}) = \frac{99}{98}$.The product becomes $\frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} \times \cdots \times \frac{99}{98}$.Notice each denominator cancels with the numerator of the next term, except the first denominator $3$ and the last numerator $99$.After cancellation, only $\frac{99}{3} = 33$ remains.Answer:33 (Option C)