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When n = 1, n! = 1! = 1 = [(n + 1) / 2)] ^ n
When n = 2, n! = 2! = 2

It is proved by mathematical induction that the nth power of 1 + 2 + 3 + 4 +... + 2 is 2N-1 power of 2 + n-1 power of 2

It is proved that (1) when n = 1, left formula = 1 + 2 ^ 1 = 3, right formula = 2 ^ (2 × 1-1) + 2 ^ (1-1) = 2 + 1 = 3
At this point, the proposition holds
(2) Suppose that when n = k, the proposition holds
1+2+3+…… +2^k=2^(2k-1)+2^(k-1)
Then when n = K + 1
1+2+3+…… +2^k+[(2^k+1)+(2^k+2)+…… +(2^k+2^k-1)+(2^k+2^k)]
=2^(2k-1)+2^(k-1)+[(2^k+1)+(2^k+2)+…… +(2^k+2^k-1)+(2^k+2^k)]
=2^(2k-1)+2^(k-1)+2^k•2^k+(1+2+3+…… +2^k)
=2^(2k-1)+2^(k-1)+2^k•2^k+2^(2k-1)+2^(k-1)
=2^(2k)+2^(k)+2^2k=2^(2k+1)+2^k
In other words, the proposition is established by mathematical induction

Using mathematical induction to prove "(n + 1) (n + 2) · •（n+n）=2n•1•3•… (2n-1) "when" n from K to K + 1 ", the left-hand side of" n from K to K + 1 "needs to be multiplied by () A. 2k+1 B. 2（2k+1） C. 2k+1 K+1 D. 2k+3 K+1

When n = k, the left end = (K + 1) (K + 2) (K + 3) （2k），
When n = K + 1, the left end = (K + 2) (K + 3) （2k）（2k+1）（2k+2），
(n + 2K + 2) is the left-hand multiplication of (2k + 1)
(K + 1) = 2 (2k + 1), so B

It is proved by mathematical induction that 1 + A + A2 + + an = 1-an + 2 / 1-A (a ≠ 1, n  n). When n = 1 is verified, the formula calculated on the left is

It's 1 + A + A ^ 2 + +A ^ n = [1-A ^ (n + 1)] / (1-A)
N = 1, left = 1 + A, right = (1-A ^ 2) / (1-A) = 1 + A, left = right
If n = k, then n = K + 1
Left = [1-A ^ (K + 1)] / (1-A) + A ^ (K + 1) = [1-A ^ (K + 1) + A ^ (K + 1) - A ^ (K + 2)] / (1-A) = [1-A ^ (K + 2)] / (1-A) = right
So the proposition holds for all positive integers!

Using mathematical induction to prove inequality 1 / (n + 1) + 1 / (n + 2) + +1/(n+n)> 13/24 Put 32.9g Na2CO3 and NaCl solid mixture into a beaker, at this time, the total mass is 202.9g, add 326.9g dilute hydrochloric acid, exactly complete reaction, after no bubbles escape, weigh again, the total mass is 525.4g. Calculate the mass fraction of solute in the solution (ignoring the dissolution of CO2) and what is the solvent in the solution after reaction

It is proved that if n = k, a = 1 / (K + 1) + 1 / (K + 2) + +If 1 / (K + k) > 13 / 24 holds, then when n = K + 1, left = 1 / (K + 2) + 1 / (K + 3) + +1 / (K + 1 + K + 1) = a + 1 / (K + 1 + k) + 1 / (K + 1 + K + 1) - 1 / (K + 1) = a + 1 / (2k + 1) - 1 / (2k + 2) = a + 1 / (2k + 1) (2k + 2) > a > 13 / 24

In the process of proving the inequality "1 / N + 1 + 1 / N + 2 + --- + 1 / 2n > 13 / 24 (n > 2, n belongs to n *) From the assumption that n = k holds to the left side of the inequality when n = K + 1 holds A adds one term 1 / 2 (K + 1) B adds two terms 1 / 2K + 1,1 / 2 (K + 1) C increases two terms 1 / 2K + 1,1 / 2 (K + 1), some decrease one term 1 / K + 1, D increase one term 1 / 2 (k + 1), and some decrease one term 1 / K + 1

If you choose C, it changes from end to end

Using the method of induction + 1 22+1 32+… +1 n2≥3n 2n+1（n∈N*）．

It is proved that: when n = 1, the conclusion holds; if n = k, the inequality holds; when n = K + 1, the left ≥ 3k2k + 1 + 1 (K + 1) 2, the following proof: 3k2k + 1 + 1 (K + 1) 2 ≥ 3 (k + 1) 2 (K + 1) + 1, make a difference to get 3k2k + 1 + 1 (K + 1) 2 − 3 (K + 1) 2 (K + 1) + 1 = K (K + 2) (K + 1) 2 (2k + 1) (2k + 3) > 0

It is proved by mathematical induction that 1 × 4 + 2 × 7 + 3 × 10 + +n（3n+1）=n（n+1）2．

It is proved that: (1) when n = 1, 3N + 1 = 4, and the left side of the equation starts from the sum of the products of 1 × 4 continuous positive integers,
Therefore, when n = 1, the left end of the equation = 1 × 4 = 4, and the right end = 4;
② Let n = k, 1 × 4 + 2 × 7 + 3 × 10 + +K (3K + 1) = K (K + 1) 2 holds,
When n = K + 1, 1 × 4 + 2 × 7 + 3 × 10 + +K (3K + 1) + (K + 1) (3K + 4) = K (K + 1) 2 + (K + 1) (3K + 4) = (K + 1) (K2 + K + 3K + 4) = (K + 1) (K + 1 + 1) 2, that is, n = K + 1
To sum up, it is concluded that + 2 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + 4 × 10 + +n（3n+1）=n（n+1）2．

It is proved by mathematical induction that: (n + 1) + (n + 2) + +（n+n）=n(3n+1) 2（n∈N*）

It is proved that: (1) when n = 1, left = 2, right = 2, the equation holds;
② If n = k, the conclusion holds, that is: (K + 1) + (K + 2) + +（k+k）=k(3k+1)
Two
When n = K + 1, the left side of the equation = (K + 2) + (K + 3) + +（k+k+1）+（k+1+k+1）=k(3k+1)
2+3k+2=(k+1)(3k+4)
Two
So when n = K + 1, the equation holds
It can be seen from ① and ② that: (n + 1) + (n + 2) + +（n+n）=n(3n+1)
2 (n ∈ n *) holds

Mathematical induction proves that 1 / N + 1 + 1 / N + 2 + 1 / N + 3 +... + 1 / 3N > 9 / 10 N > = 2 1) When n = 2, left = 1 / 3 + 1 / 4 + 1 / 5 + 1 / 6 = 57 / 60 > 54 / 60 = 9 / 10 (2) Suppose n = k, there is 1 / (K + 1) + 1 / (K + 2) +... + 1 / 3K > 9 / 10 Then 1 / (K + 2) + 1 / (K + 3) +... + 1 / 3 (K + 1) =[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1) >9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1) =9/10 That is, the proposition holds when n = K + 1, Thus, the original inequality holds for n ∈ N and N > 1 Why in the second step >9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1) It shouldn't be >9 / 10 + 1 / (3K + 1) + 1 / (3K + 2) + 1 / (3K + 3) - 1 / (K + 1)

1) If n = 2, left = 1 / 3 + 1 / 4 + 1 / 5 + 1 / 6 = 57 / 60 > 54 / 60 = 9 / 10, it holds. (2) if n = k, there is 1 / (K + 1) + 1 / (K + 2) +... + 1 / 3K > 9 / 10, then 1 / (K + 2) + 1 / (K + 3) +... + 1 / 3 (K + 1) = [1 / (K + 1) + 1 / (K + 2) +... + 1 / 3 (K + 1) = [1 / (K + 1) + 1 / (3K + 1) + 1 / (3K + 2) + 1 / (3K + 3) - 1 / (...)