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