Question

show that n square+n is divisible by 2 for every positive integer n​

Answer 1

Answer: to show n2+n IS DIVISIBLE BY 2

USING EUCLID DIVISION LEMMA

A=BQ+R

HERE WE HAVE TWO CASE IF N IS ODDD

OR N IS EVEN

CASE1

IF N IS EVEN

N2 +N TAKING N AS COMMON

N(N+1) IF N IS EVEN ie2,4 6,8 10 any thing it is divided by 2

CASE2

IF N ISS ODD

THEN N(N+1) HERE IF N IS ODD THEN N+1 IS ALWAYS EVEN

EG 3+1,5+1,7+1

OK SO N+1 IS DIVISIBLE BY 2 HENCE

N(N+1) IS DIVISIBLE

OR N2+ N

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Step-by-step explanation:

Answer 2

Answer:

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