# Consecutive integers...

• April 10th 2012, 05:40 PM
Wilmer
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?
• April 10th 2012, 09:04 PM
princeps
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 $c=9999$ , last integer = $14998$

in terms of $c$ , last integer = $\frac{3c-1}{2}$
• April 11th 2012, 02:29 AM
Wilmer
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.
• April 11th 2012, 02:50 AM
princeps
Re: Consecutive integers...
$a_i=\frac{c-1}{2}+i ~\text{ for }~ 1\leq i \leq c$

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