The value of \( A \) is derived from the expression \( A = (1 - \frac{1}{(1+2)}) \times (1 - \frac{1}{(1+2+3)}) \times \ldots \times (1 - \frac{1}{(1+2+\ldots+2013)}) \). The calculations involve evaluating the cumulative sums in the denominators, leading to a precise computation of \( A \). The final answer confirms the correctness of the derived value, showcasing the effectiveness of the approach used in the discussion.
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Albert said:$A=(1-\dfrac{1}{(1+2)})\times(1-\dfrac{1}{(1+2+3)})\times---\times(1-\dfrac{1}{(1+2+---+2013)})$
please find the value of $A$
great ! your answer is correctkaliprasad said:This is a product of $t_k$ for k = 1 to 2012 where
$t_k= (1-\dfrac{1}{(1+2+ \cdots + k+1)})$
= $( 1- \dfrac{2}{(k+1)(k+2)})$
= $\dfrac{k(k+3)}{(k+1)(k+2)}$
so the product = $\dfrac{1* 2015}{3 * 2013}= \dfrac{2015}{6039}$