1. Formula method:
Sum formula of arithmetic sequence: sn = n (a1+an)/2 = na1+n (n-1) d/2.
Sum formula of proportional series: Sn=na 1(q= 1)
sn = a 1( 1-qn)/( 1-q)=(a 1-an×q)/( 1-q)(q≠ 1)
2. Dislocation subtraction
Applicable question type: it is suitable for the form of a series in which the general term formula is an arithmetic linear function and multiplied by an equal proportion.
{an} and {bn} are arithmetic progression and geometric progression respectively. sn = a 1b 1+a2b2+a3b3+...+anbn。
For example:
an=a 1+(n- 1)d
bn=a 1？ q(n- 1)
Cn=anbn
TN = a 1b 1+a2 B2+a3 B3+a4 B4....+anbn
qTn= a 1b2+a2b3+a3b4+...+a(n- 1)bn+anb(n+ 1)
TN-qTn = a 1b 1+B2(a2-a 1)+B3(a3-a2)+...bn[an-a(n- 1)]-anb(n+ 1)
TN( 1-q)= a 1b 1-anb(n+ 1)+d(B2+B3+B4+...bn)
=a 1b 1-an？ b 1？ qn+d？ B2[ 1-q(n- 1)]/( 1-q)
Tn= above formula /( 1-q)
3. Reverse addition
This is a method to derive the sum formula of the first n terms of arithmetic progression, that is, arrange a series in reverse order (reverse order) and then add it to the original series to get n (a 1+an).
Sn = a1+A2+A3+...+An
Sn =an+ a(n- 1)+a(n-3)......+a 1
Add up and down to get 2Sn, that is, Sn= (a 1+an)n/2.
4. Grouping method
There is a series, which is neither arithmetic progression nor geometric progression. If this kind of series is properly decomposed, it can be divided into several arithmetic progression, proportional series or ordinary series, and then summed and merged separately.
For example: an=2n+n- 1
5. Split terminology method
General term formula applicable to fractional form. A term is divided into two or more difference forms, that is, an = f (n+ 1)-f (n), and then many intermediate terms are cancelled when accumulating.
Commonly used formula:
( 1) 1/n(n+ 1)= 1/n- 1/(n+ 1)
(2) 1/(2n- 1)(2n+ 1)= 1/2[ 1/(2n- 1)- 1/(2n+ 1)]
(3) 1/n(n+ 1)(n+2)= 1/2[ 1/n(n+ 1)- 1/(n+ 1)(n+2)]
(4) 1/(√a+√b)=[ 1/(a-b)](√a-√b)
(5) n？ n！ =(n+ 1)！ -No!
[Example] Find the sum of the first n items in a series an= 1/n(n+ 1).
Solution: an =1/n (n+1) =1/n-1/(n+1) (split term)
rule
tin
= 1- 1/2+ 1/2- 1/3+ 1/4 ...+ 1/n- 1/(n+ 1)
= 1- 1/(n+ 1)
= n/(n+ 1)
Summary: This deformation is characterized by the fact that after each item in the original series is split into two items, most of the items in the middle cancel each other out. There are only a few things left.
Note: The remaining projects have the following characteristics.
1 The position of the remaining items is symmetrical before and after.
The positive and negative of the other items are opposite.
6. Mathematical induction
Generally speaking, to prove a proposition related to a positive integer n, there are the following steps:
(1) proves that the proposition holds when n takes the first value;
(2) Assuming that n = k (the first value of k ≥ n, and k is a natural number), it is proved that the proposition is also true when n=k+ 1.
Example:
Verification:
1×2×3×4+2×3×4×5+3×4×5×6+……+n(n+ 1)(n+2)(n+3)=[n(n+ 1)(n+2)(n+3)(n+4)]/5
Prove:
When n= 1, there are:
1×2×3×4 + 2×3×4×5 = 2×3×4×5×( 1/5 + 1) = 2×3×4×5×6/5
Assuming that the proposition holds when n=k, then:
1×2×3×4+2×3×4×5+3×4×5×6+……+k(k+ 1)(k+2)(k+3)=[k(k+ 1)(k+2)(k+3)(k+4)]/5
Then when n=k+ 1, there is:
1×2×3×4+2×3×4×5+3×4×5×6+……+(k+ 1)(k+2)(k+3)(k+4)
= 1×2×3×4+2×3×4 * 5+3×4×5×6+……+k(k+ 1)(k+2)(k+3)+(k+ 1)(k+2)(k+3)(k+4)
=[k(k+ 1)(k+2)(k+3)(k+4)]/5+(k+ 1)(k+2)(k+3)(k+4)
=(k+ 1)(k+2)(k+3)(k+4)*(k/5+ 1)
=[(k+ 1)(k+2)(k+3)(k+4)(k+5)]/5
That is, when n=k+ 1, the original equation still holds, which is proved by induction.
Step 7 generalize
Simplify the general formula first, and then sum.
For example, find the sum of the first n items of a sequence 1, 1+2, 1+2+3, 1+2+3+4, ... At this time, calculate an first, and then sum by grouping and other methods.
8. Joint total:
For example:1-2+3-4+5-6+...+(2n-1)-2n.
Method 1: (merge)
Find the sum of odd and even terms, and then subtract them.
Method 2:
( 1-2)+(3-4)+(5-6)+……+[(2n- 1)-2n]