1. ## Consecutive integers...

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?

2. ## Re: Consecutive integers...

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;
for $\displaystyle c=9999$ , last integer =$\displaystyle 14998$

in terms of $\displaystyle c$ , last integer =$\displaystyle \frac{3c-1}{2}$

3. ## Re: Consecutive integers...

Good one Princeps.
Accidentally got that while fooling around with consecutive numbers.
Had you seen the "easy formula" last integer = (3c - 1)/2 before?
Amazing that there is a solution for ALL odd c's.

4. ## Re: Consecutive integers...

$\displaystyle a_i=\frac{c-1}{2}+i ~\text{ for }~ 1\leq i \leq c$

$\displaystyle \displaystyle \sum_{i=1}^c \left(\frac{c-1}{2}+i\right)=c\cdot\frac{(c-1)}{2}+\displaystyle \sum_{i=1}^c i=\frac{c(c-1)}{2}+\frac{c(c+1)}{2}=c^2$