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 = $\dfrac{1* 2015}{3 * 2013}= \dfrac{2015}{6039}$
The equation is A = 2013/2014.
To solve for A, you must first calculate the sum of the series 1+2+...+2013. This can be done by using the formula n(n+1)/2, where n is the last number in the series (in this case, 2013). This gives us a sum of 2013*2014/2 = 2013*1007 = 2,021,091.
Next, we substitute this sum into the original equation, giving us 1 - 1/2,021,091 = A. Simplifying, we get A = 2013/2,021,091 = 0.0009941.
This equation is known as a telescoping series, where many terms cancel out during the calculation, leaving a simpler expression. It is also an example of a partial sum of a harmonic series, which is a series where the terms decrease in size but never reach zero. This equation also has practical applications in engineering and physics, such as in calculating the total resistance of a series circuit.
Yes, this equation can be generalized for any positive integer n. The sum of the series 1+2+...+n can be expressed as n(n+1)/2, so the equation for solving for A would be A = n/(n+1).
Aside from its applications in engineering and physics, this equation can also be used in finance and economics. For example, it can be used to calculate the present value of an annuity, where A represents the periodic payments and n represents the number of periods. It can also be used in probability calculations, such as finding the probability of rolling a certain number on a dice after a certain number of rolls.