Consider a sequence where the $$n^{th}$$ term, $$t_n = n/(n+2), n =1, 2, ....$$ The value of $$t_3 * t_4 * t_5 * …..* t_{53}$$ equals.
substituting 3,4...53 in the given function, we get
$$t_3 = \frac{3}{5}$$
$$t_4 = \frac{4}{6}$$
$$t_5 = \frac{5}{7}$$
$$t_6 = \frac{6}{8}$$
Multiplying the values, we get $$\frac{3}{5}*\frac{4}{6}*\frac{5}{7}*....\frac{52}{54}*\frac{53}{55} $$ which ultimately after cancellations give $$\frac{3*4}{54*55}=\frac{2}{495}$$
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