Consider a sequence where the nth term, \( t_n = \frac{n}{n+2} \), \( n = 1, 2, \dots \). The value of \( t_3 \times t_4 \times t_5 \times \dots \times t_{53} \) equals:
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In a telescoping product \( \frac{n}{n+k} \), the number of terms remaining at the start and end is equal to the difference \( k \). Here \( k=2 \), so two terms remain on top and two on the bottom.