Proof: for any positive integer n, (2n + 1) & sup2; - 1 must be divisible by 8

Proof: for any positive integer n, (2n + 1) & sup2; - 1 must be divisible by 8

（2n＋1）^2－1＝［（2n＋1）＋1］［（2n＋1）－1］＝4n（n＋1）.
∵ n is a positive integer, ∵ n, (n + 1) are two adjacent positive integers, and one of ∵ n, (n + 1) must be even,
N (n + 1) is even, 4N (n + 1) can be divided by 8

（2-3×5-1）+（4-3×5-2）+… （2n-3×5-n）=______ ．

The original formula = (2 + 4 +...) +2n）-3×（5-1+5-2+… +5-N) = n (2 + 2n) 2-3 × 5 − 1 (1 − 5 − n) 1 − 5 − 1 = n (n + 1) - 3 × 1 − 5 − N4 = n (n + 1) - 34 [1 - (15) n]

sn=(2-3x5^-1)+(4-3x5^-2)+… +(2n-3x5 ^ - n) to explain the answer in detail, it's urgent!

sn=(2-3x5^-1)+(4-3x5^-2)+… +(2n-3x5^-n)
=（2+4+.+2n)-(3x5^-1+3x5^-2+.+3x5^-n)
=（2^1+2^2+.+2^n）-(3x5^-1+3x5^-2+.+3x5^-n)
=2x(1-2^n)/（1-2）+3x5^-1x(1-0.2^n)/（1-0.2）
It's full