Question

Prove that product of two consecutive integers is positive when divided by 2

Answer 1

$\huge{\bf{\underline{\red{Solution :}}}}$

⇒Suppose the two consecutive positive integers be z and (z +1)

Now According to Question :

⇒Products of two consecutive positive integers = z(z +1) = $\bold{z^{2} + z}$

$\bf{\boxed{\bold{ Case - i) \ z \ is \ even \ number }}}$

Suppose z = 2k

$\implies \bold{ z^{2} + z }$

$\implies \bold{ (2k)^{2} + 2k}$

$\implies \bold{4k^{2} + 2k}$

$\implies \bold{2k(2k + 1)}$

Therefore , the product is divisible by 2

$\bf{\bold{\red{Case - ii) \ z \ is \ odd \ number}}}$

Suppose z = 2k + 1

$\implies \bold{z^{2} + z }$

$\implies \bold{(2k + 1)^{2} + (2k + 1)}$

$\implies \bold{4k^{2} + 4k + 1 + 2k + 1}$

$\implies \bold{4k^{2} + 6k + 2}$

$\implies \bold{2(2k^{2} + 3k + 1)}$

Therefore , the product is divisible by 2

( From both Condition the proved that the product of two consecutive integers is positive when divided by 2)

$\rule{200}2$

Answer 2

SOLUTION

Let the 2 consecutive positive integer be x and x+1.

Let, x=2a(even) and x+1 =2a+1 (odd)

Product of 2 interesting (x) (x+1)

= x² + x

CASE 1:

➡If x is an even number .

= x² + x = 2a²+2a

$= 2a(a + 1)$

Check:

➡divide the above expression by 2

$2a(a + 1) \2$

WE GET,

$= a(a + 1)$

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CASE 2:

➡If x is an odd number

By putting values x= 2a+1

WE HAVE ,

x²+ x = (2a+1) ² + (2a+1)

4a²+1+4a+2a+1

2(2a² + 3a + 1)

Check :

➡Divide the above expression by 2

2(2a² + 3a + 1)²

2a² + 3a + 1

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HENCE VERIFIED.

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