|
Question 1209812: Simplify \frac{1 + 3 + 5 + ... + 1999 + 2001 + 2003}{2 + 4 + 6 + ... + 2000 + 2002 + 2004 + 2006 + 2008 + 2010}.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $N = 1 + 3 + 5 + \dots + 2003$ and $D = 2 + 4 + 6 + \dots + 2010$.
We want to find $\frac{N}{D}$.
The numerator $N$ is the sum of an arithmetic progression with first term $a_1 = 1$, last term $a_n = 2003$, and common difference $d = 2$.
To find the number of terms $n$, we use the formula $a_n = a_1 + (n-1)d$:
$2003 = 1 + (n-1)2$
$2002 = 2(n-1)$
$1001 = n-1$
$n = 1002$
The sum of an arithmetic series is given by $S_n = \frac{n(a_1 + a_n)}{2}$.
So, $N = \frac{1002(1+2003)}{2} = \frac{1002(2004)}{2} = 1002(1002) = 1004004$.
The denominator $D$ is the sum of an arithmetic progression with first term $b_1 = 2$, last term $b_m = 2010$, and common difference $d = 2$.
To find the number of terms $m$, we use the formula $b_m = b_1 + (m-1)d$:
$2010 = 2 + (m-1)2$
$2008 = 2(m-1)$
$1004 = m-1$
$m = 1005$
The sum of an arithmetic series is given by $S_m = \frac{m(b_1 + b_m)}{2}$.
So, $D = \frac{1005(2+2010)}{2} = \frac{1005(2012)}{2} = 1005(1006) = 1011030$.
Then $\frac{N}{D} = \frac{1004004}{1011030}$.
We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor.
Let's divide both by 6:
$\frac{1004004}{6} = 167334$
$\frac{1011030}{6} = 168505$
So $\frac{N}{D} = \frac{167334}{168505}$.
Let's check if there are any more common factors.
We can use the Euclidean algorithm to find the gcd of 167334 and 168505.
$168505 = 1(167334) + 1171$
$167334 = 142(1171) + 832$
$1171 = 1(832) + 339$
$832 = 2(339) + 154$
$339 = 2(154) + 31$
$154 = 4(31) + 30$
$31 = 1(30) + 1$
The gcd is 1, so the fraction is already in its simplest form.
Thus, $\frac{N}{D} = \frac{1004004}{1011030} = \frac{167334}{168505}$.
Final Answer: The final answer is $\boxed{\frac{167334}{168505}}$
|
|
|
| |
