How to prove this inequality by mathematical induction? N ^ 3 < when n is greater than or equal to 6 Please give a complete answer! Brother There is a passage that I didn't understand. How did I get it? (k+1)!-(k+1)^3 >(k+1)k^3-k^3-3k^2-3k-1

How to prove this inequality by mathematical induction? N ^ 3 < when n is greater than or equal to 6 Please give a complete answer! Brother There is a passage that I didn't understand. How did I get it? (k+1)!-(k+1)^3 >(k+1)k^3-k^3-3k^2-3k-1

Prove: when n = 6, 6 ^ 3 = 2166! = 7206 ^ 3 (K + 1) k ^ 3-K ^ 3-3k ^ 2-3k-1 = k ^ 4-3k ^ 2-3k-1 = (k ^ 4-1) - 3K (K + 1) = (k ^ 2 + 1) (k-1) - 3K (K + 1) = (K + 1) [(k ^ 2 + 1) (k-1) - 3K] = (K + 1) (k ^ 3-K ^ 2 + k-1-3k) = (K + 1) (k ^ 3-K ^ 2-2k-1) = (K + 1) (k ^ 3-2k ^ 2 + k ^ 2-2k-1) = (K + 1) = (K + 1) = (k ^ 3-2k ^ 2-2k-1) = (K + 1) = (k ^ 3-2k ^ 2-2k-1) = (K + 1)

Using mathematical induction to prove inequality
1/(n+1）+1/(n+2)+1/(n+3)+… +1/3n＞5/6 (n≥2）

(1) when n = 2, the original formula = 1 / 3 + 1 / 4 + 1 / 5 + 1 / 6 = 57 / 60 > 5 / 62.) suppose that when n = k, (k is any number greater than 2) there exists 1 / (K + 1) + 1 / (K + 2) + 1 / (K + 3) + +So, when n = K + 1, the original formula is 1 / (K + 2) + 1 / (K + 3) + 1 / (K + 4) + +1/3k+1/(3k+1)+1/(3k+2)+1/(3k+3)...

How to prove inequality by mathematical induction
The square of n is less than the nth power of 2

Certification:
(1) When n = 0,1,2,3, 2 ^ n > n ^ 2 holds
(2) Suppose 2 ^ k > k ^ 2 holds when n = K (k > = 3)
When n = K + 1, 2 ^ (K + 1) = 2 * 2 ^ k = 2 ^ k + 2 ^ k > k ^ 2 + K ^ 2
And K ^ 2-2k-1 = k ^ 2-2k + 1-2 = (k-1) ^ 2-2, k > = 3, k-1 > = 2, (k-1) ^ 2 > = 4,
(k-1) ^ 2-2 > = 2 > 0, so K ^ 2-2k-1 > 0, K ^ 2 > 2K + 1, K ^ 2 + K ^ 2 > k ^ 2 + 2K + 1 = (K + 1) ^ 2
SO 2 ^ (K + 1) > (K + 1) ^ 2, that is, when n = K + 1, 2 ^ n > n ^ 2 also holds
In conclusion, when n is a natural number, the square of n is less than the nth power of 2