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