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