Mathematical induction is one of the key contents of college entrance examination. Analogy and conjecture are prominent ideas, abstractions and generalizations in inductive method of applied mathematics, and they are a main applied thinking method from special to general.

● Difficult magnetic field

(★★★★★) Is there a, b and c that make the equation 1? 22+2? 32+…+n(n+ 1)2= (an2+bn+c)。

● Case study

It is proved that whether the positive numbers A, B and C are arithmetic progression or geometric progression, when n > 1, n ∈ n * is not equal to A, B and C, there are: an+CN > 2bn.

Proposition intention: This question mainly investigates the inequality proved by mathematical induction, which belongs to the class of ★★★★★.

Relying on knowledge: the nature of arithmetic progression and geometric progression and the general steps of proving inequality by mathematical induction.

Wrong solution analysis: inequality should be proved to be true for both geometric series and arithmetic series, not just one case.

Skills and methods: the conclusion used in this question is: (AK-CK) (a-c) > 0 holds (a, b and c are positive numbers), so AK+ 1+CK+ 1 > AK? c+ck？ a.

It is proved that (1) let a, b and c be geometric series, and A =, C = BQ (Q > 0 and q≠ 1).

∴ an+cn =+bnqn = bn (+qn) > 2 billion

(2) let a, b and c be arithmetic progression, then 2b=a+c conjecture > () n (n ≥ 2 and n∈N*).

The following is proved by mathematical induction:

(1) When n=2, it is 2 (A2+C2) > (A+C) 2, ⅶ.

(2) Let n=k, i.e.

Then when n=k+ 1, (AK+1+CK+1+AK+1+CK+/)

>(ak+ 1+ck+ 1+ak？ c+ck？ a)= (ak+ck)(a+c)

>()k？ ()=( )k+ 1

[Example 2] In the sequence {an}, a 1= 1. When n≥2, an, sn and sn- become geometric series.

(1) Find A2, A3 and A4, and derive the expression of an;

(2) prove the conclusion by mathematical induction;

(3) Find the sum of all items in the sequence {an}.

Proposition intention: This topic examines the basic knowledge of sequence, mathematical induction, sequence limit and so on.

The essence of geometric series and the general steps of mathematical induction. The methods used are induction, guessing and proof.

Wrong solution analysis: in (2), sk =- should be abandoned, which is often ignored.

Skills and methods: Finding the general term can prove that {} is a arithmetic progression with {} as the first term, with tolerance, and then get the general term formula.

Solution: ∫an, Sn, sn- path geometric series,? (Sn- )(n≥2) (*)

(1) is derived from A 1 = 1, S2 = A 1+A2 = 1+A2, and substituted into the formula (*): A2 =-

Substitute a 1= 1, A2 =-, S3 =+A3 into the formula (*): A3 =-

Similarly: A4 =-, from which it can be deduced that: an=

(2) (1) When n= 1, 2, 3, 4, (*) shows that the conjecture holds.

② If n=k(k≥2), AK =- holds.

So sk2 =-? (Sk-)

∴(2k-3)(2k- 1)sk2+2sk- 1=0

∴Sk= (shed)

By Sk+ 12=ak+ 1？ (sk+ 1-)，(sk+AK+ 1)2 = AK+ 1(AK+ 1+sk-)

From ① ②, an= holds for all n ∈ n 。

(3) The sum of the first n items in the sequence of (2) is Sn= ,∴S= Sn=0.

● Tips

The basic form of (1) mathematical induction

Let P(n) be a proposition about natural number n, if

1 P (N0) established (laid the foundation)

2 assuming that P(k) holds (k≥n0), it can be deduced that P(k+ 1) holds (induction), then for all natural numbers n ≥ n0, P(n) holds.

(2) the application of mathematical induction

Specific proofs of common mathematical induction: identities, inequalities, divisibility of numbers, calculation problems in geometry, sum of general terms of series, etc.

● Destroy difficult training.

First, multiple choice questions

1.(★★★★) Is f(n)=(2n+7) known? 3n+9 has a natural number m, so that for any n∈N, m can be divisible by f(n), and the maximum value of m is ().

A.30 B.26 C.36 D.6

2. (★★★★★) Prove 3k≥n3(n≥3, n∈N) by mathematical induction. The first step should be to verify ().

A.n= 1 B.n=2 C.n=3 D.n=4

Second, fill in the blanks

3. (★★★★★★) Observe the following formula: ... and then _ _ _ _ _ _ _.

4. (★★★★★) If A 1 = and AN+ 1 = are known, the values of A2, A3, A4 and A5 are _ _ _ _ _ _ _ _ _ _ _ _ _ _.

Third, answer questions.

5. (★★★★★★★) Prove by mathematical induction that 4 +3n+2 is divisible by 13, where n∈N*.

6. (★★★★★) If n is a natural number greater than 1, verify:.

7. (★★★★★★) The known series {bn} is arithmetic progression, B 1 = 1, b1+B2+…+b10 =145.

(1) Find the general formula BN of the sequence {bn};

(2) Let the general term an of the sequence {an} be loga (1+) (where a > 0 and a≠ 1) and let Sn be the sum of the first n terms of the sequence {an}. Try to compare the sizes of Sn and logabn+ 1 to prove your conclusion.

8. (★★★★★★) Let the real number q satisfy | q | < 1 and the sequence {an} satisfy: a 1=2, a2≠0, an? An+ 1 =-qn, find an expression, and if S2n < 3, find the range of q. 。

Reference answer

Hard magnetic field

Solution: suppose a, b and c make the equation hold, then let n= 1, 2,3, right.

Therefore, for n= 1, 2, 3, the following equation holds.

1? 22+2? 32+…+n(n+ 1)2=

Remember Sn= 1 22+2? 32+…+n(n+ 1)2

Let n=k, that is, sk = (3k2+11k+10).

Then sk+1= sk+(k+1) (k+2) 2 = (k+2) (3k+5)+(k+1) (k+2) 2.

= (3k2+5k+ 12k+24)

=〔3(k+ 1)2+ 1 1(k+ 1)+ 10〕

In other words, this equation also applies to n=k+ 1.

To sum up, when a = 3, b = 1 1 and c = 10, it is assumed that all natural numbers n hold.

Destroy difficult training

I. 1. Analysis: ∫f( 1)= 36, f (2) = 108 = 3× 36, f (3) = 360 = 10× 36.

∴f( 1),f(2),f(3) are divisible by 36, and f(n) is divisible by 36.

Proof: when n= 1, 2, it is proved from above, and when n=k(k≥2),

f(k)=(2k+7)？ 3k+9 is divisible by 36, so when n=k+ 1,

f(k+ 1)-f(k)=(2k+9)？ 3k+ 1？ -(2k+7)？ 3k

=(6k+27)？ 3k-(2k+7)？ 3k

=(4k+20)？ 3k=36(k+5)？ 3k-2？ (k≥2)

F(k+ 1) is divisible by 36.

∵f( 1) cannot be divisible by a number greater than 36, and the maximum m value is equal to 36.

Answer: c

2. analysis: from the meaning of the question, n≥3 and ∴ n=3 should be verified.

Answer: c

Second, three. Analysis:

(n∈N*)

(n∈N*)

、 、 、

3.5. Prove: (1) When n= 1, 42×1+1= 91is divisible by 13.

(2) Assuming that 42k+ 1+3k+2 can be divisible by 13 when n=k+1,

42(k+ 1)+ 1+3k+3 = 42k+ 1？ 42+3k+2？ 3-42k+ 1？ 3+42k+ 1？ three

=42k+ 1？ 13+3? (42k+ 1+3k+2？ )

∫42k+ 1？ 13 is divisible by 13, and 42k+ 1+3k+2 is divisible by 13.

It also holds when n=k+ 1

According to ① ②, when n∈N*, 42n+ 1+3n+2 is divisible by 13.

6. Prove: (1) When n=2,

(2) suppose it holds when n=k, that is

7.( 1) solution: let the tolerance of the sequence {bn} be d, and from the meaning of the question, ∴ bn = 3n-2.

(2) Prove that BN = 3N-2 shows that

sn = loga( 1+ 1)+loga( 1+)+…+loga( 1+)

= loga〔 1+ 1〕( 1+)……( 1+)〕

And logabn+ 1=loga, how about Sn and logabn+ 1? Compare the size of (1+1) (1+) ... (1+) with.

Let n= 1 and use (1+ 1)= 1

Let n=2 and (1+ 1)( 1+)

Speculation: (1+1) (1+) ... (1+) > (*)

① When n= 1, the formula (*) is verified.

② If n=k(k≥ 1), the formula (*) holds, that is, (1+1) (1+) >.

Then when n=k+ 1,

That is, when n=k+ 1, the formula (*) holds.

From ① ②, it can be seen that formula (*) holds for any positive integer n. 。

Therefore, when a > 1, Sn> logabn+ 1+0? , when 0 < a < 1, Sn< logabn+ 1+0?

8. solution: ∵a 1? a2=-q，a 1=2，a2≠0，

∴q≠0,a2=-，

∵ Ann? an+ 1=-qn，an+ 1？ an+2=-qn+ 1？

Divide the two formulas and you get, an+2=q? One; one

So, a 1=2 and a3=2? q，a5=2？ Qn… conjecture: A2n+ 1 =-Qn (n = 1, 2,3, …)

Comprehensive ① ②. Guess that the general formula is an=

The following proof: (1) The conjecture holds when n= 1, 2.

(2) When n = 2k- 1, A2k- 1 = 2? When qk- 1 is n=2k+ 1, a2k+ 1=q? a2k- 1？

∴a2k+ 1=2？ Qk is n = 2k- 1.

It can be inferred that n=2k+ 1 also holds.

If n=2k and a2k=- qk, then when n=2k+2, a2k+2=q? a2k？ ,

So A2K+2 =-QK+ 1, which shows that n=2k holds, and it can be inferred that n=2k+2 also holds.

To sum up, this conjecture holds for all natural numbers n.

The general formula thus obtained is an=

s2n =(a 1+a3…+a2n- 1)+(a2+a4+…+a2n)

=2( 1+q+q2+…+qn- 1？ )- (q+q2+…+qn)

Because | q | < 1, ∴ =

Know < 3, and pay attention to 1-q > 0, while | q | < 1 gets-1 < q < 0 or 0 < q < 1.

One, two and three points that should be paid attention to in the proof of mathematical induction.

As the basic method of proving mathematical propositions, mathematical induction can complete the proof of many propositions related to natural numbers. Of course, any method has its limitations, and mathematical induction is no exception.

Verification example:.

Evidence 1: Remember.

∵ , , , ,

Sometimes it is true. The following evidence (time)

(1) when, obviously established;

② Assumption () holds, that is,

Then.

and

That is, when it was established.

By ① and ②, the sum () holds.

This () is established.

Method 2: Remember,

Then,

Because the value in each bracket is positive, it holds. When again,

.

When.

So it's true for me.

First, it can be seen from the above proof that not all propositions about natural numbers must be proved by mathematical induction; In many cases, it is simpler and more convenient to prove by other methods than mathematical induction.

Example 1 Verification: ()

Prove that 1: ① It is timely and valid.

(2) Assuming the conclusion is true, then when,

=

That is, when the proposition holds. Therefore, it is established.

Method 2: ∵, ∴ If you want to prove it, you only need to prove that ∵ ∴ is established.

∴( 1860) established.

Proof 3: Left = Right.

Left