2. The college entrance examination is very flexible now, so it is generally not taken. The following are its proof steps for your reference.

We use concrete examples to illustrate.

Example 1 observe the following two series, from which one do you get that an is always less than bn? Prove your conclusion.

{an=n2}: 1，4，9， 16，25，36，49，64，8 1，…；

{bn=2n}:2，4，8， 16，32，64， 128，256，5 12，…；

Analysis: From the first few conjectures of the sequence, from the fifth item, An, that is, N2.

When proving the above conjecture by mathematical induction, step (1) should prove the case of n = 5.

Proof: (1) When n = 5, there are 52

(2) Assuming that the proposition holds when n=k(k≥5), there is K2.

When n=k+ 1, because (k+1) 2 = k2+2k+1< k2+3k.

& ltk2+k2 = 2 k2 & lt； 2×2k=2k+ 1

So (k+ 1) 2

According to (1)(2), N2

Example 2 proves Bernoulli inequality:

If x is a real number and x >；; -1, x≠0, n is a natural number greater than 1, then there is (1+x) n >; 1+nx。

Analysis: Bernoulli inequality involves two letters, X represents any real number greater than-1 and not equal to 0, and N is a natural number greater than 1. We can only generalize n by mathematical induction.

Prove: (1) When n=2, (1+x) 2 =1+2x+x2 > 1+2x, the inequality holds.

(2) Assuming that the inequality holds when n=k(k≥2), there is (1+x) k >; 1+kx。

When n=k+ 1, (1+x) k+1= (1+x) (1+x) k > (1+x)

= 1+x+kx+kx2 & gt； 1+(k+ 1)x。

So when n=k+ 1, the inequality holds.

According to (1)(2), Bernoulli inequality holds.

Reflection: Bernoulli inequality can be used for numerical estimation and scale method to prove inequality. For example, when x is a real number and x >; When-1, x≠0, it is not difficult to get the inequality [1-x/(1+x)] n ≥1-NX/(1+x) from Bernoulli inequality, which is not less than.

Explanation: It is difficult to prove inequality by mathematical induction, and it often appears that the proposition holds when free n=k, and it holds when deducing free n=k+ 1. In order to complete this proof, we should not only use the inductive hypothesis correctly, but also use the relevant knowledge of other conditions and problems flexibly.

There is another situation, please see the following example:

Example 3: Test: All natural numbers n(n≥ 1) have 2n+2 >；; n2。

Analysis: When n= 1, the inequality is obviously established. By convention, the next step should be 2k+2 >；; When the inequality is proved again, K2 is also true for n=k+ 1, but since 2k+1+2-(k+1) 2 = 2 (2k+2)-k2-2k-3 >: 2× k2-k2-2k-3 = (.

If there can be (k- 1)2-4≥0, the inequality will be established, but there must be a condition of k≥3, so we have to move the starting point to n=3, so we can deduce it by inductive hypothesis. Although the starting point has been improved, it is easy to be summarized.

Example 4 Verification:1+1/22+1/32+…+1/N2

Analysis: If n=k, the inequality holds, that is,1+1/22+1/32+…+1/k2.

It is proved that 1+ 1/22 = 5/4 when n=2.

The conclusion holds if n=k, that is,1+1/22+1/32+…+1/k2.

That is, when n=k+ 1, the conclusion holds, which is obtained by mathematical induction.

1+ 1/22+ 1/32+…+ 1/N2 & lt； 2- 1/n，

Thus1+1/22+1/32+…+1/N2 < 2 (n ≥1).

Reflection: General propositions provide a stronger inductive hypothesis, which will be easier to prove by mathematical induction. Therefore, "actively strengthening propositions" is indeed a skill worth pondering.