Simplify (1+3+5+...+199)/(2+4+6+...+200).

Multiply 100 to both the numerator and denominator.

$\begin{array}{rl}&\dfrac{1}{1.01}\\=&\dfrac{100}{101}\end{array}$

1 + 3 + 5 + ..... + 199   can be expressed as the sum n^2  for n terms

The number of terms in this sum =  [ 1 + 199] / 2  = 200 / 2 = 100

So

1 + 3 + 5 + ....+199  = n^2 =  100^2  

And

 2 + 4 + 6 + ....+ 200    can be expressed  as the sum   n(n +1) for n terms

The number of terms in this sum  is the same as the first  = 100

So .... 2 + 4 + 6 + ....+ 200  =  n (n + 1)  = 100(101)  

So  

[1 + 3 + 5 + ....+199 ] / [ 2 + 4 + 6 + ....+ 200]  =  [ 100^2] / [ 100 * 101]  =

[100 * 100] / [ 100 * 101] =  1 / 101

CPhill: How do you get: [100 * 100] / [ 100 * 101] =  1 / 101?

It should be: 1/1.01. No??!!.

Sorry for the slight error....of course the answer should be  100/101 = 1/1.01  = 0.9900990099009901

Simplify (1+3+5+...+199)/(2+4+6+...+200).

$\begin{array}{|rcll|} \hline s_n &=& \frac{(a_1+a_n)}{2} \cdot n \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \frac{1+3+5+...+199}{2+4+6+...+200} \\\\ &=& \frac{\frac{(a_1+a_n)}{2} \cdot n }{\frac{(b_1+b_n)}{2} \cdot n} \\\\ &=& \frac{a_1+a_n}{b_1+b_n}\\\\ &=& \frac{1+199}{2+200}\\\\ &=& \frac{200}{202}\\\\ &=& \frac{100}{101}\\\\ &=& 0.99009900990\dots \\ \hline \end{array}$