MHB Solving for A: 1 - 1/(1+2+...+2013)

[Albert1]

$A=(1-\dfrac{1}{(1+2)})\times(1-\dfrac{1}{(1+2+3)})\times---\times(1-\dfrac{1}{(1+2+---+2013)})$
please find the value of $A$

 

[kaliprasad]

$A=(1-\dfrac{1}{(1+2)})\times(1-\dfrac{1}{(1+2+3)})\times---\times(1-\dfrac{1}{(1+2+---+2013)})$
please find the value of $A$

Spoiler

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 = $\dfrac{1* 2015}{3 * 2013}= \dfrac{2015}{6039}$

 

[Albert1]

Spoiler

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 = $\dfrac{1* 2015}{3 * 2013}= \dfrac{2015}{6039}$

great ! your answer is correct