Math, asked by Avi2019, 11 months ago

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Q
show that the product of two consecutive positive integers is divisible by 2​

Answers

Answered by Anonymous
12

Answer:

Let the two consecutive positive integers be x and x+1

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

According to the given question

Product of two integers (x)(x+1)

=x^{2}+x=x

2

+x

Case 1: if x is an even number

By putting the values

We have x^{2}+x=2 a^{2}+2 ax

2

+x=2a

2

+2a

=2a (a+1)

Check and divide the above expression by 2

= 2a (a+1)/2

We get a (a+1)

Hence it is clearly proved that the product of 2 integers are divisible by 2

Case 2: If x is an odd number

By putting the values x=2a+1

We have x^{2}+x=(2 a+1)^{2}+(2 a+1)x

2

+x=(2a+1)

2

+(2a+1)

\begin{lgathered}\begin{array}{l}{=4 a^{2}+1+4 a+2 a+1} \\ {=4 a^{2}+6 a+2} \\ {=2\left(2 a^{2}+3 a+1\right)}\end{array}\end{lgathered}

=4a

2

+1+4a+2a+1

=4a

2

+6a+2

=2(2a

2

+3a+1)

Check: Divide the above expression by 2

=2\left(2 a^{2}+3 a+1\right) / 2=2(2a

2

+3a+1)/2

We get 2 a^{2}+3 a+12a

2

+3a+1

Hence proved

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