Type one
Inductive conjecture proof
We can write the first few terms of the sequence from the recursive formula of the sequence, then sum up the rules from the first few terms, guess a general term formula of the sequence, and finally prove it by mathematical induction.
Type 2
"Difference Law" and "Product Commercial Law"
(1) When the recurrence formula of a series can be changed into an+ 1-an=f(n), take n = 1, 2, 3, ..., n- 1, and get n- 1 formulas.
a2-a 1=f( 1)，a3-a2=f(2)，…，an-an- 1=f(n- 1)，
And if f (1)+f (2)+…+f (n-1) can be found, the general term an can be accumulated on both sides. This method is called "step by step method".
(2) When the recurrence formula of a series can be changed into an+ 1/an=f(n), let n= 1, 2,3, …, n- 1, then we can get n- 1 formulas, that is.
A2/A 1 = F ( 1)，A3/A2 = F (2)，A4/A3 = F (3)，…，an/An- 1 = F (n- 1)，F ( 1)
Type 3
structured approach
The recurrence formula is pan=qan- 1+f(n)(p and q are nonzero constants), and a new geometric series solution can be constructed by using the undetermined coefficient method.
Type 4
Can be converted into type three passthrough items.
(1) "Logarithmic method" was transformed into type III.
The recurrence formula is an+ 1=qan? K (q > 0, k ≠ 0 and k ≠ 1, a 1 > 0), take the common logarithm on both sides and get lgan+ 1=klgan+lgq. Let lgan=bn, then we have bn+/kloc-.
(2) Convert "Reciprocal Method" into Type 3.
The recurrence formula is in the form of quotient: an+1= (pan+b)/(qan+c) (an ≠ 0, PQ ≠ 0, PC ≠ QB).
If b=0, an+ 1 = pan/(qan+c). Because an≠0, the reciprocal of both sides is 1/an+ 1=q/p+c/pan, so bn =1/an.
If b≠0, let an+ 1+x=y(an+x)/qan+c, compare with the known recurrence formula to get x and y, let bn=an+x, and get bn+ 1=ybn/qan+c, which is converted into b=0.
Type five
The recurrence formula is an+ 1/an=qn/n+k(q≠0, k∈N).
We can multiply the two sides of the equation (n+k) an+ 1=qnan by (n+k-1) (n+k-2) ... (n+1) to get (n+k-/kloc-). Nan, then bn+1= (n+k) (n+k-1) (n+k-2) ... (n+1) an+1.
So bn+ 1=qbn, so the sequence {bn} is the common ratio q, and the first term is b1= k (k-1) (k-2) ... 1? a 1=k！ The geometric series of a 1, and then an can be obtained.
In a word, the problem of finding the general term formula from the recursive formula of sequence is complicated and it is impossible to discuss them one by one. But as long as we grasp the recursive relationship of recursive sequence, analyze the structural characteristics and be good at reasonable deformation, we can find an effective way to solve the problem.
Type one? Inductive conjecture proof
We can write the first few terms of the sequence from the recursive formula of the sequence, then sum up the rules from the first few terms, guess a general term formula of the sequence, and finally prove it by mathematical induction.
Example 1? Let the series {an} be a positive series with the first term 1, (n+1) a2n+1-nan2+an+1an = 0 (n =1,2, 3, ...
Solution: decompose (n+1) a2n+1-nan2+an+1an = 0 (n =1,2,3, ...) into (an+1+an).
Because an > 0, (n+ 1)an+ 1=nan, that is, an+1= n/(n+1) an.
Therefore, A2 = (1/2) a1= (1/2), A3 = (2/3) A2 = (1/3), ... Guess an = (1/).
Type two? "Difference Law" and "Product Commercial Law"
(1) When the recurrence formula of a series can be changed into an+ 1-an=f(n), take n = 1, 2, 3, ..., n- 1, and get n- 1 formulas.
a2-a 1=f( 1)，a3-a2=f(2)，…，an-an- 1=f(n- 1)，
And if f (1)+f (2)+…+f (n-1) can be found, the general term an can be accumulated on both sides. This method is called "step by step method".
Example 2? It is known that the sequence {an} satisfies a 1 = 1, an = 3n- 1+an- 1 (n ≥ 2). It is proved that an = (3n- 1)/2.
(Question 19 in the 2003 National Mathematics Volume)
It is proved that from the known an-an- 1=3n- 1, therefore,
an =(an-an- 1)+(an- 1-an-2)+…+(a2-a 1)+a 1 = 3n- 1+3n-2？ +…+3+ 1=3n- 1/2。
So I got the certificate.
(2) When the recurrence formula of a series can be changed into an+ 1/an=f(n), let n= 1, 2,3, …, n- 1, then we can get n- 1 formulas, that is.
a2/a 1=f( 1)，a3/a2=f(2)，a4/a3=f(3)，…，an？ /an- 1？ =f(n- 1)？ ,? If f (1) f (2) f (3) ... you can get f (n- 1), and you can get an by multiplying the two sides. This method is called product management method.
Example 3? (The same example is1) (Question 15 of National Mathematics Volume in 2000)
Another solution: (n+1) a2n+1-nan2+an+1an = 0 (n? = 1, 2, 3, ...) and get (n+ 1)an+ 1=nan, that is.
an+ 1/an=n/(n+ 1)。 ?
So an=an/an- 1? an- 1/an-2？ An -2/ An -3? …? a2/a 1？ =n- 1/n？ n-2/n- 1？ n-3/n-2？ … ? 1/2? = 1/n。
Type three? structured approach
The recurrence formula is pan=qan- 1+f(n)(p and q are nonzero constants), and a new geometric series solution can be constructed by using the undetermined coefficient method.
Example 4? (Same Example 2) (Question 19 of the 2003 National Mathematics Volume)
Another solution: 3 is obtained from an=3n- 1+an- 1 an/3n = an- 1/3n- 1+ 1。
Let bn = an/3n, then there is
bn = 1/3bn- 1+ 1/3。 (*)
Let bn+x= 1/3(bn- 1+x), then bn =1/3bn-1+/3x-x, compared with formula (*), x =-60. Therefore, bn-1/2 =1/3 (bn-1-1/2). Therefore, the sequence {BN- 1/2} is the first term of b1-1= a. (1/3) n-1,namely an/3n-1/2. Therefore, an = 3n [65438+]
Example 5? In the sequence {an}, a 1 = 1, an+ 1 = 4an+3n+ 1, find an.
Solution: let an+1+(n+1) x+y = 4 (an+NX+y), then
An+ 1 = 4an+3x+3y-x, compared with the known an+ 1=4an+3n+ 1, we get
3x=3, so
x= 1，
3y-x= 1，y=(2/3)。
So the sequence {an+n+(2/3)} is a geometric series with the first term a 1+ 1+(2/3)=(8/3), so an+n+(2/3)=(8/3)? 4n- 1, i.e.
Ann =(8/3)? 4n- 1-n-(2/3)。
Another solution: when n≥2 is known, an = 4an-1+3 (n-1)+1,which is different from the known relationship, an+1-an = 4 (an-an-1. Therefore, the sequence {an+ 1-an+ 1} is a geometric series with the first term a2-a1+1= 8, and the general term an can be obtained by "differential method". 4n- 1-n-(2/3)。
Type four? Can be converted into
Type 3 general terms
(1) "logarithmic method" is converted into
The third kind.
The recurrence formula is an+ 1=qan? K (q > 0, k ≠ 0 and k ≠ 1, a 1 > 0), take the common logarithm on both sides and get lgan+ 1=klgan+lgq. Let lgan=bn, then we have bn+/kloc-.
The third kind.
Example 6? Given the sequence {an}, a 1=2, an+ 1=an2, find an.
Solution: from an+ 1 = AN2 > 0, the logarithm of both sides is lgan+ 1 = 2lgan. Let bn=lgan, then bn+ 1 = 2bn. Therefore, the first term of the sequence {bn} is b 1 = lga 1 =.
(2) Convert the "reciprocity method" into
The third kind.
The recurrence formula is in the form of quotient: an+1= (pan+b)/(qan+c) (an ≠ 0, PQ ≠ 0, PC ≠ QB).
If b=0, an+ 1 = pan/(qan+c). Because an≠0, the reciprocal of both sides is 1/an+ 1=q/p+c/pan, so bn =1/an.
The third kind.
If b≠0, let an+ 1+x=y(an+x)/qan+c, compare with the known recurrence formula to get x and y, let bn=an+x, and get bn+ 1=ybn/qan+c, which is converted into b=0.
Example 7? In the sequence {an}, a 1 = 2, an+1= (3an+1)/(an+3) are known, and the general term an is found.
Solution: Let an+ 1+x=y(an+x)/an+3, then an+1= (y-x) an+(y-3) x/an+3, combined with the known recurrence formula.
Y-x=3, so
x= 1，
y-3= 1，y=4，
Then an+1+1= 4 (an+1)/an+3, and bn=an+ 1, then bn+ 1=4bn/bn+2, and the reciprocal is/kloc-0. 1/bn+ 1/4, that is,1/bn+1-1/2 =1/2 (1/bn-/kloc-0.
Therefore, the first term of the sequence {1/BN- 1/2} is1/b1-1/2 =1/a1-65438.
So1/bn-1/2 = (-1/6) (1/2) n-1,so you can get an. Finding the sum of the first n items of a series is one of the teaching focuses of the chapter "Series" in senior high school mathematics. However, for some non-arithmetic progression, as long as we seriously explore the characteristics of these series. And its structure is not irregular.
Typical example:
1, using the general formula:
Law: You can use the general term formula to write the terms of series, so that sum can be recombined into the sum of series.
Example 1: Find the sum of the first n terms of 5, 55, 555, ….
Solution: ∫ an = 5 9 (10n-1)
∴sn = 5 ^ 9( 10- 1)+5 ^ 9( 102- 1)+5 ^ 9( 103- 1)+…+5 ^ 9( 10n- 1)
= 5 9[( 10+ 102+ 103+…+ 10n)-n]
=( 10n+ 1-9n- 10)
2, dislocation subtraction:
General terrain such as {an? The sequence of bn}, where {an} is arithmetic progression and {bn} is geometric progression, can be summed by dislocation subtraction.
Example 2: find: sn =1+5x+9x2++(4n-3) xn-1.
Solution: Sn =1+5x+9x2++(4n-3) xn-1①.
(1) both sides are multiplied by x, so.
x Sn=x+5 x2+9x3++(4n-3)xn ②
①-②，( 1-x)sn = 1+4(x+x2+x3 ++)-(4n-3)xn。
When x= 1, sn =1+5+9++(4n-3) = 2n2-n.
When x≠ 1, Sn =1-x [4x (1-xn)1-x+1-(4n-3) xn].
3, crack offset method:
The characteristic of this kind of series is that each term of the series is the product of two or more adjacent terms in arithmetic progression.
Usually, {an} is an arithmetic series with a tolerance of d, then:
That is to say, the split term cancellation method is mainly used to sum the fractional series whose denominator is the product of adjacent k terms in arithmetic progression and whose numerator is constant. In order to summarize the splitting term elimination method, the splitting term can be determined by the undetermined coefficient method.
Example 3: Sum 1 3, 1 1 5, 1 3 5, 1 63.
Solution:
4. Grouping method:
In some series, two or more arithmetic progression or geometric progression can be obtained by proper grouping, and the sum of the original series can be obtained by using the sum formula of arithmetic progression or equal ratio series.
Example 4: Find the sum of the first n items of a series.
Solution:
5. Polymerization method:
Some series are more complicated, and each item is the sum of several numbers, so the aggregation method is often adopted.
First sum the nth term, and then simplify the general term, thus changing the form of the original sequence and facilitating the solution.
Example 5: Find the sum of the first n terms of the sequence 2, 2+4, 2+4+6, 2+4+6+8, …, 2+4+6+…+2n, ….
Solution: ∫ an = 2+4+6+…+2n = n (n+1) = N2+n.
∴sn=( 12+ 1)+(22+2)+(32+3)+……+(N2+n)
=( 12+22+32+…+N2)+(+2+3+…+n)
= n(n+ 1)(2n+ 1)+n(n+ 1)
= 1 3n(n+ 1)(n+2)
6. Reverse addition:
The derivation of arithmetic progression's first N-term summation formula is to arrange the items in the summation formula in reverse order to get another summation formula, and then add it to the original summation formula accordingly to get arithmetic progression's first N-term summation formula. This method can also be used to find the sum of the first n terms of some series.
Example 6: Given lg(xy)=a, find S, where
S=
Solution: arrange the items in sum S in reverse order, and get
Add this summation formula and the original summation formula on both sides, and you get
2S= + +？ +
Project (n+ 1)
=n(n+ 1)lg(xy)
∫LG(xy)= a∴s = n(n+ 1)a
Although the above six methods have their own characteristics, the general principle is to be good at changing the formal structure of the original series, so that it can be eliminated or solved by using the summation formula of arithmetic progression or equal ratio series and other known basic summation formulas. As long as we grasp this law, the summation of series can become difficult.