Question

Prove that ${n^2} + n$ is divisible by 2 for any positive integer n.

Answer 1

Hint: In this question we have to solve two cases. In the first case we have to put n=2q, where “q” is some integer and solve for a result which is multiple of 2 or not. Similarly, solve for n=2q+1 and find if the result is multiple of 2 or not.

Complete step by step solution:
 We have,
 ${n^2} + n$ --- eq.1
We know that any integer is in the form of 2q or 2q+1, where q is some integer.
When , n = 2q,
On putting the value of n equal to 2q in eq.1, we get
$ \Rightarrow {n^2} + n = {\text{ (2q}}{{\text{)}}^2} + 2{\text{q}}$
On simplifying above expression, we get
$
  {\text{ = 4}}{{\text{q}}^2} + 2{\text{q}} \\
  {\text{ = 2q(2q + 1)}} \\
  {\text{ }} \\
 $
Above equation , ${\text{2q(2q + 1)}}$ is divisible by 2. --- Statement 1
Now, let n = 2q+1
$ \Rightarrow {n^2} + n = {\text{ (2q + 1}}{{\text{)}}^2} + (2{\text{q + 1)}}$
On simplifying above expression, we get
$
  {\text{ = 4}}{{\text{q}}^2} + 4{\text{q}} + 1 + 2{\text{q + 1}} \\
  {\text{ = 4}}{{\text{q}}^2} + 6{\text{q + 2}} \\
  {\text{ = 2(2}}{{\text{q}}^2} + 3{\text{q + 1)}} \\
 $
Above equation, ${\text{2(2}}{{\text{q}}^2} + 3{\text{q + 1)}}$ is divisible by 2.-- Statement 2
Therefore, from statement 1 and 2 we can conclude that ${n^2} + n$ is divisible by 2 for any positive integer n.

Note: Whenever you get this type of question the key concept to solve is to learn the concept of numbers which are divisible by 2. A number is divisible by 2 if the digit at unit place is either 0 or multiple of 2. So a number is divisible by 2 if the digit at its unit place is 0,2,4,6,8. Numbers divisible by 2 are called even numbers. And the numbers not divisible by 2 are called odd numbers.