Mathematical induction to prove inequality (1 / N + 1) + (1 / N + 2) +. + (1 / 3N + 1) > 25 / 24

Mathematical induction to prove inequality (1 / N + 1) + (1 / N + 2) +. + (1 / 3N + 1) > 25 / 24

It is proved that when n = 1, the inequality is tenable by 1 / 2 + 1 / 3 + 1 / 4 = 13 / 12 = 26 / 24 > 25 / 24. Now let n = k, the inequality holds, that is, 1 / (K + 1) + 1 / (K + 2) +... + 1 / (3K + 1) > 25 / 24

Prove 1 + 2 + 3 + by mathematical induction +n2=n4+n2 2, then when n = K + 1, the left end should add () A. k2+1 B. （k+1）2 C. (k+1)4+(k+1)2 Two D. （k2+1）+（k2+2）+（k2+3）+… +（k+1）2

When n = k, the left end of the equation = 1 + 2 + +k2，
When n = K + 1, the left end of the equation = 1 + 2 + +k2+（k2+1）+（k2+2）+（k2+3）+… +(K + 1) 2, the term 2K + 1 is added
Therefore, D

The second step is to prove that from K to K + 1, the number of items added at the left end is () A 2^(k-1) B 2^k -1 C 2^k D 2^k +1

When n = k, left = 1 + 1 / 2 + 1 / 3 +... + 1 / 2 ^ (k-1)
When n = K + 1, left = 1 + 1 / 2 + 1 / 3 +... + 1 / 2 ^ K
From K to K + 1, the number of items added on the left is 2 ^ K-2 ^ (k-1) = 2 ^ (k-1)
Choose B

Using mathematical induction to prove inequality: 1 N+1 n+1+1 n+2+… +1 N ∈ n ＞ 1

It is proved that: (1) when n = 2, left = 1
2+1
3+1
4＝13
12 ＞ 1,  n = 2 (2 points)
(2) If n = K (K ≥ 2), it holds
One
K+1
k+1+1
k+2+… +1
k2＞1
So when n = K + 1, left = 1
k+1+1
k+2+1
k+3+… +1
(k+1)2
=1
K+1
k+1+1
k+2+1
k+3+… +1
k2+2k+1
(k+1)2−1
K
＞1+1
k2+1+1
k2+2+… +1
(k+1)2−1
K
＞1+(2k+1)•1
(k+1)2−1
k＞1+k2−k−1
k2+2k+1＞1
ν n = K + 1 also holds (7 points)
According to (1) (2), the inequality obtained holds for all n > 1 (8 points)

It is known that (1 + 1 / x) ^ x is infinitely close to y = e when x > = 1. It is proved by mathematical induction that the inequality (n / 3) ^ n < < (n / 2) ^ n when n > = 6 The problem is difficult, so give 100 points

Is the problem difficult?
As long as we know 2 (n / 3) ^ n * (n + 1) = [n / (n + 1)] ^ n * [(n + 1) / 3] ^ n * (n + 1) > 1 / 3 * [(n + 1) / 3] ^ n * (n + 1) = [(n + 1) / 3] ^ {n + 1}
(n+1)!= n!(n+1) < (n/2)^n * (n+1) = [n/(n+1)]^n * [(n+1)/2]^n * (n+1) < 1/2 * [(n+1)/2]^n * (n+1) = [(n+1)/2]^{n+1}

By means of mathematical induction, the inequality 1 / (2 ^ 2) + 1 / (3 ^ 2) + for any positive integer n greater than 1 is proved +1 / (n ^ 2) less than (n-1) / n

1) If n = 2, 1 / 2 ^ 2 = 1 / 4 = 2), then, for n = K + 1, there are
1/2^2+a/3^2+…… +1/k^2+1/(k+1)^2

How to prove 1 + 2 + 3 +... + n = 1 / 2n (n + 1) by mathematical induction

How to prove 1 + 2 + 3 +... + n = 1 / 2n (n + 1) by mathematical induction
Proof: when n = 1, left = 1, right = 1,2 * 1 (1 + 1) = 1, left = right;
Let n = k, the equation holds, that is: 1 + 2 + 3 +... + k = 1 / 2K (K + 1);
When n = K + 1,
Left = 1 + 2 + 3 +.. + K + (K + 1)
=[1 + (K + 1)] + [2 + k] + [3 + (k-1)] +... [there are 1 / 2 (K + 1) items in total]
=(2 + k) + (2 + k) + (2 + k) +... [there are 1 / 2 (K + 1) items in total]
=1 / 2 (K + 1) (K + 2) = right
The proof is over

Using mathematical induction to prove Equality: n ∈ n, n ≥ 1, 1 − 1 2+1 3−1 4+… +1 2n−1−1 2n＝1 n+1+1 n+2+… +1 2n．

It is proved that: (1) when n = 1, left = 1 − 1
2＝1
2 = right, the equation holds
(2) Suppose that the equation holds when n = K,
That is, 1 − 1
2+1
3−1
4+… +1
2k−1−1
2k＝1
k+1+1
k+2+… +1
2K
Then 1 − 1
2+1
3−1
4+… +1
2k−1−1
2k+(1
2k+1−1
2k+2)=1
k+1+1
k+2+… +1
2k+(1
2k+1−1
2k+2)=1
k+2+… +1
2k+1
2k+1+1
When n = K + 1, the equation also holds
Synthesis (1) (2), equality holds for all positive integers

The three sides a, B, C of triangle ABC satisfy the equation a + B + C = AB + BC + AC. please judge the shape of ABC and explain the reason

A + B + C = AB + BC + AC 2A + 2B + 2C = 2Ab + 2BC + 2Ac a-2ab + B + c-2ca + A + b-2bc + C = 0 (a-b) + (B-C) + (C-A) = 0 A-B = 0 a = B B B-C = 0 B a = b = b = C equilateral triangle

It is known that the triangle ABC is 3ABC, and the box xpyz is the Y power of - 4x and the Z power of P. the speed of Mn3 multiplied by nm25 is urgent

Hello:
Triangle ABC = 3ABC, box xpyz = - 4x ^ y * P ^ Z
Triangle Mn3 times box nm25
=(3*m*n*3)*(-4n^2*m^5)
=-36m^6n^3
Hope to help you!