Question

Prove that (n? +n) is divisible by 2 for any positive integer n.
[CBSE 2​

Answer 1

Answer:

Step-by-step explanation:

n^2+n = n(n+1)

n can only be even or odd

case 1 : n is even

even(even +1)

now even +1 =odd

even(odd) =even

case2 n is odd

odd(odd+1)

odd+1 =even

odd(even) =even

all even no.s are divisble by 2 so n^2+n is divisble by 2 as it is always even

Answer 2

Casei: Let n be an even positive integer. When n = 2q In this case , we have n2 - n = (2q)2 - 2q = 4q2 - 2q = 2q (2q - 1 ) n2 - n = 2r , where r = q (2q - 1) n2 - n is divisible by 2 . Case ii: Let n be an odd positive integer. When n = 2q + 1 In this case n2 -n = (2q + 1)2 - (2q + 1)= (2q +1) ( 2q+1 -1)= 2q (2q + 1) n2 - n = 2r , where r = q (2q + 1) n2 - n is divisible by 2. ∴ n 2 - n is divisible by 2 for every integer n