Show that for any rational number n, the expression 2n ∙ 4^(n+1) ∙1/8^n is equal to 4.

answer:

step-by-step explanation:

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

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry