Prove that for any natural n the value of the expression (n+7)^2-n^2 is divisible by 7 ( )·7
∧
Put the answer as parentheses

65

step-by-step explanation:

37

step-by-step explanation:

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube