Sum of 2 consecutive integers is $x$ then their product will be?

Question

According to a source, the answer is $x^2-\frac{1}{4}$

Please explain

Your source has a typo. It should be $\frac{x^2 - 1}4$ rather than $x^2 - \frac 14$ — fleablood, Jul 21 '16 at 08:33
Think about it. Suppose the consecutive numbers are n and m. Then x = n + m. This is an integer. $n*m$ is also an integer. As is $x^2$ But $x^2 - 1/4$ is never an integer. — fleablood, Jul 21 '16 at 09:05

score 2 · Answer 1 · answered Jul 21 '16 at 08:18

2

Let $n$ be the first of those consecutive numbers. Then we have: $$ n + (n+1) = x $$

which implies $ n= \frac{x-1}{2}$

using that you can calculate the product of $n(n+1)$

answered Jul 21 '16 at 08:18

Snufsan

According to this, the answer would be $\frac{x^2-1}{4}$ – Usama36 Jul 21 '16 at 08:21
Yep. Someone at the source made the all to common editing error of writing $\frac{x^2 - 1}{4}=x^2 - 1/4 = x^2 - \frac 14$. So your source has a typo. – fleablood Jul 21 '16 at 08:32
@fleablood It doesn´t seem that your guess is right. – callculus42 Jul 21 '16 at 08:52
What guess? I didn't guess at anything. The solution is wrong. Take 7+8 = 15. 7 x 8 = 56 = (225 - 1)/4 $\ne$ 255 - 1/4. – fleablood Jul 21 '16 at 09:01

score 0 · Accepted Answer · answered Jul 21 '16 at 08:29

0

Let the consecutive integers be $a$,$a+1$...

Now,$a+(a+1)=x\implies a=\frac{x-1}{2}\implies (a+1)=\frac{x+1}{2}$

Hence, $a(a+1)=\frac{x-1}{2}*\frac{x+1}{2}=\frac{x^2-1}{4}$

answered Jul 21 '16 at 08:29

COOLGUY

2 Answers2