+0  
 
0
3
2
avatar+16 

When expressed as a common fraction, what is the value of (2 + 4 + 6 + 8 + ... + 1340 + 1342)/(3 + 6 + 9 + 12 + ... + 2010 + 2013)?

 Dec 20, 2023
 #1
avatar+222 
+1

First, we recognize that both the numerator and denominator are arithmetic series.

 

The numerator is an arithmetic series with first term a1​=2, common difference d=2, and last term an =1342.

The denominator is an arithmetic series with first term a1​=3, common difference d=3, and last term an​=2013.

 

To find the number of terms in each series, we can use the formula for the $n$th term of an arithmetic series: an​=a1​+(n−1)d.

 

For the numerator: 1342=2+(n−1)(2)⇒n=671.

 

For the denominator: 2013=3+(n−1)(3)⇒n=671.

 

Now, we can use the formula for the sum of an arithmetic series: Sn​=2n​(a1​+an​).

 

The sum of the numerator is Sn​=2671​(2+1342)=450546.

 

The sum of the denominator is Sn​=2671​(3+2013)=673546.

 

Therefore, the value of the fraction is 450546/673546​.

 

We can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2. So, the final answer is 225273/336773​​.

 Dec 20, 2023
 #2
avatar+126665 
+1

We can write

 

2 ( 1 + 2 + 3 + 4 +......+ 670 + 671)                2

____________________________   =       ____

3 ( 1 + 2 + 3 + 4 +.......+670 + 671)                3

 

 

cool cool cool

 Dec 21, 2023

2 Online Users

avatar