Can you show that (n + 7) & sup2; - (N-5) & sup2; is divisible by 24

Can you show that (n + 7) & sup2; - (N-5) & sup2; is divisible by 24

(n+7)²-(n-5)²
= [ (n+7)+(n-5) ] * [(n+7)-(n-5)]
= (2n + 2) * 12
= 24(n+1)
With a factor of 24, it can be divided by 24

Given 1 + 2 + 3 +... + n = M. find the algebraic formula (a ^ n by B) (a ^ n-1 by B ^ 2) (a ^ n-2 by B ^ 3) The value of (AB ^ n)

(a ^ n by B) (a ^ n-1 by B ^ 2) (a ^ n-2 by B ^ 3) (ab^n)
=[a^(n+n-1+n-2+...+1)]*b^(1+2+3+...+n)
=[a^m]*b^m
=(ab)^m

It is known that 3 ^ m = A and 3 ^ = B are expressed as 3 ^ - M-N by the algebraic formula of AB respectively

Hello, the original formula = 1 / 3 ^ (M + n)
=1/(3^m×3^n)
=1/(ab)