The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from r to s, inclusive, where 0<r<s?
Choose the correct choice from the following answers
A) s(s+1)/2 - (r+1)(r+2)/2	B) s(s+1)/2 - r(r+1)/2	C) s(s+1)/2 - (r-1)r/2	D) (s-1)s/2 - (r+1)(r+2)/2	E) (s-1)s/2 - r(r+1)/2
Or try plug-in method: let s=4 and r=3 --&gt; then s+r=7. Let see which option yields 7.   A. s(s+1)/2 - (r+1)(r+2)/2 = 10-10= 
<gadget id="calculator">10-10</gadget>
<output>0</output>
0;   B. s(s+1)/2 - r(r+1)/2= 10-6= 
<gadget id="calculator">10-6</gadget>
<output>4</output>
4;   C. s(s+1)/2 - (r-1)r/2= 10-3= 
<gadget id="calculator">10-3</gadget>
<output>7</output>
7 --&gt; OK;   D. (s-1)s/2 - (r+1)(r+2)/2= 6-10=-4;   E. (s-1)s/2 - r(r+1)/2 = 6-6= 
<gadget id="calculator">6-6</gadget>
<output>0</output>
0.   Answer: s(s+1)/2 - (r-1)r/2.
<result>C</result>
The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from r to s, inclusive, where 0<r<s?