The sum of c consecutive positive integers = c^2.
Example, c=5: 3+4+5+6+7 = 25 ; 5^2 = 25
(last integer = 7)
If c = 9999, what is the last integer?
What is the last integer in terms of c?
for $\displaystyle c=9999$ , last integer =$\displaystyle 14998$Code:c:=9999: for n from 1 to 10000 do if sum(k,k=n..n+c-1)=c^2 then print(n+c-1); break; end if; end do;
in terms of $\displaystyle c$ , last integer =$\displaystyle \frac{3c-1}{2}$
$\displaystyle a_i=\frac{c-1}{2}+i ~\text{ for }~ 1\leq i \leq c$
$\displaystyle \displaystyle \sum_{i=1}^c \left(\frac{c-1}{2}+i\right)=c\cdot\frac{(c-1)}{2}+\displaystyle \sum_{i=1}^c i=\frac{c(c-1)}{2}+\frac{c(c+1)}{2}=c^2$