Prove that for every natural number n, n^2+n is even.?

Answer 1

Case i
n is odd
=>n^2 is odd
=>n^2 + n is even ( odd num + odd num is even )
Case ii
Follow same principles.

Or even better
n^2 + n = n(n+1)
=>It is the product of two consecutive numbers
=>At least one is an even number
=>Product is even

Answer 2

let n=3
therefore n^2+n=3^2+3
=9+3
=12
let n=4
therefore n^2+n=4^2+4
=16+4
=20
hence every natural number n is even for the above case

Answer 3

If you are want to prove it using mathematical induction:

Prove statement true for n:=1
(1)^2+1:=2
Statement is true for n:=1

Assume statement true for n:=k
i.e (k)^2+k gives an even number

Prove statement true for n:=k+1
i.e (k+1)^2+k+1

(k+1)^2+K+1=k^2+2k+1+K+1
=((k)^2+k)+2k+2

Statement is true as the result will give an even number.

By the principle of mathematical induction the statement is true for every natural number n.

Answer 4

First factor it to get n(n+1)

n and n+1 are consecutive numbers so one will be even and one odd

The product of an even and odd number is always even

Answer 5

Simple: if n is odd, then n + 1 is even, so their product

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

is even. Similarly, if n is even, then n^2 + n is again the product of an odd and an even number, and so is even.

Answer 6

n^2+n=n(n+1) is the procut of two consecutive numbers so one is even and so the product

Answer 7

n^2 + n = n(n + 1) so that if n is even then (n + 1 is odd and the product od odd and even = even

If n is odd then (n + 1) is even ans again the product of even and odd = even

Answer 8

a)P(a million) - 3*a million/= ok^2+6>3k+3 for this reason ok^2+4+2k >3k+3