Question

1. Prove that for any natural value of n the value of the expression:

a) (n+1)^2–(n–1)^2 is divisible by 4
b) (2n+3)^2–(2n–1)^2 is divisible by 8

Answer 1

(a)

Use the fact that
`a^2-b^2=(a+b)(a-b)`. In this case a=n+1, b=n-1.

I am using "4|expression" to denote that 4 should divide the expression:

`4|[(n+1)^2-(n-1)^2]\implies 4|(n+1+n-1)(n+1-n+1)\implies\\\implies4|2n\cdot2\implies 4|4n \,\,\,\mbox{true for all }n\in N`

(b)

Same trick

`8|[(2n+3)^2-(2n-1)^2]\implies 8|(2n+3+2n-1)(2n+3-2n+1)\implies\\\implies 8|(2n+2)\cdot4 \implies 8|8(n+1)\,\,\,\mbox{true for all } n\in N`

Answer 2

Explanation:

sorry if i made small errors

a)

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

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

n^2+2n+1 - (n^2-2n+1)

expand

= n^2+2n+1 - n^2+2n-1

simplify

= 4n is divisible by 4

answer = 4(n) or 4 times n

b)

(2n+3)(2n+3) = 4n^2+12n+9

(2n-1)(2n-1) = 4n^2-4n+1

4n^2+12n+9 - (4n^2-4n+1)

expand

= 4n^2+12n+9 - 4n^2+4n-1

simplify

= 16n+8 is divisible by 8

answer = 8(2n+1)