It is proved that the value of 5N power minus one of 2 can be divisible by 31

It is proved that the value of 5N power minus one of 2 can be divisible by 31

Because 2 ^ 5n-1 = (2 ^ 5) ^ n - (1) ^ n = (2 ^ 5-1) [(2 ^ 5) ^ (n-1) + +1^(n-1)]=31*[(2^5)^(n-1)+…… +1^(n-1)]
So it's divisible by 31
If you don't understand, please ask

Prove 1 + 1 / 2 + 1 / 3 +... + 1 / N > ln (n + 1) + n / 2 (n + 1)
Don't use constructors, use definite integrals

Refer to the following figure: ① & nbsp; ln (n + 1) = & nbsp; ∫ {1, N + 1} 1 / X DX is the area between X = 1 and x = n + 1 above the x-axis under the curve y = 1 / X. ② & nbsp; 1 + 1 / 2 + 1 / 3 +... + 1 / N is the sum of the areas of N rectangles, which completely contains the area in ①

Prove (1 + 2 + 3 + n) (1 + 1 / 2 + 1 / 3 +. 1 / N) ≥ N2 + n-1 by mathematical induction
For all positive integers n greater than 2, the following inequality holds
(1 + 2 + 3 +... + n) (1 + 1 / 2 + 1 / 3 +...... 1 / N) ≥ the square of N + n-1

(1)n=3
Left = (1 + 2 + 3) (1 + 1 / 2 + 1 / 3) = 6 * (1 + 1 / 2 + 1 / 3) = 6 + 3 + 2 = 11
Right = 3 * 3 + 3-1 = 11
Therefore, the inequality holds when n = 3
(2) If n = K (K ≥ 3), the inequality holds
That is, (1 + 2 + 3 +. + k) (1 + 1 / 2 +. + 1 / k) ≥ K & # 178; + k-1
When n = K + 1,
Left = [1 + 2 + 3 +. + K + 1 / (K + 1)] * [1 + 1 / 2 +. + 1 / K + 1 / (K + 1)]
=(1+2+3+.+k)(1+1/2+.+1/k)+(1+2+3+.+k)*(1/k+1)+(k+1)*[1+1/2+.+1/k+1/(k+1)]
≥k²+k-1+k(k+1)/2* (1/k+1)+(k+1)[1+1/2+1/(k+1)]
=k²+k-1+k/2+k+1+(k+1)/2+1
>k²+k-1+k/2+k+1+k/2+1
=k²+3k+1
=(k+1)²+(k+1)-1
So when n = K + 1, the inequality holds
So the inequality holds for all positive integers n greater than 2

12+22+32+42…… +(n-1) 2 + N2 =? Is the square, the square of one plus the square of two is added to n

=n(n+1)(2n+1)/6