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