Foreword
Scope
① When the number of columns often contain \ ((- 1) ^ k \) or \ ((- 1) ^ { k + 1} \) symbols such as number of columns, their entries are often reflected Positive and Negative interval occurs, this time is often often two item (or three, etc.) into a group and then summed;
② When the number of columns often contain \ ((- 1) ^ k \) or \ ((- 1) ^ { k + 1} \) , etc. When the number of symbols columns, can also consider the number of columns that grouped into odd-numbered columns and even-numbered number of columns, and then summing the packet;
③ When the number of columns contained in \ (a_n + a_ {n + 1} = f (n) \) form, or \ (a_n + a_ {n + 1} + a_ {n + 2} = f (n) \) the form can be considered and sUM.
Related formula
① arithmetic progression \ (S_n = \ cfrac {n (a_1 + a_n)} {2} = na_1 + \ cfrac {n (n-1) \ cdot d} {2} \)
② geometric series \ (S_n = \ left \ { \ begin {array} {l} {na_1, q = 1} \\ {\ cfrac {a_1 \ cdot (1-q ^ n)} {1-q} = \ cfrac {a_1-a_nq} {1-q}, q \ neq 1} \ end {array} \ right. \)
③\(1+2+3+\cdots+ n=\cfrac{n(n+1)}{2}\)；
④ \ (. 1. 5 + + +. 3 \ cdots + (. 1-2N) = \ cfrac {[1+ (. 1-2N)] \ n-CDOT} ^ {n-2} = 2 \) , note that the number of summation term \ (\ n) term;
⑤ \ (. 4 + 2 + +. 6 \ cdots = 2N + \ cfrac {(2N + 2) \ n-CDOT} ^ {n-2} = 2 \) , note that the number of summation term \ (n-\) item;
⑥\(1^2+2^2+3^2+\cdots+ n^2=\cfrac{n\cdot (n+1)\cdot (2n+1)}{6}\)；
⑦\(1^3+2^3+3^3+\cdots+ n^3=[\cfrac{n(n+1)}{2}]^2\)；
⑧ by the \ (a_ {n + 2} -a_n = 2 \) can be seen, the number of columns in the odd-numbered arithmetic progression, tolerance \ (2 \) ; even term arithmetic progression, tolerance \ (2 \) ;
⑨ by the \ (\ cfrac {a_ {n + 2}} {a_n} = 2 \) can be seen, as the number of columns in the odd-numbered geometric, geometric progression \ (2 \) ; even term as geometric, geometric progression \ (2 \) ;
Computing skills
① Index calculation:
② using the arithmetic progression seeking entry columns:
The (D A_N = A_1 + (n--. 1) \ CDOT \) \ , available several \ (n = \ cfrac {A_N-A_1} {D} + 1'd \) , promotion to give a number of \ (n = \ cfrac A_N-a_m {} {D} m + \) ,
Shown with column \ (2 ^ 1,2 ^ 3,2 ^ 5, \ cdots, 2 ^ {2n-1} \) calculated number of items, the number of items which can be used to calculate the standard, which is marked as just like the difference between the number of columns,
项数\ (r = \ {cfrac a_1-a_n} {+ d} = 1 \ {cfrac (2n-1) -1} {3-1} = n + 1 \) ;
Typical Example Analysis
Method 1: items and the summation method, since the title has \ (n-\) th power, so that for \ (n-\) partial parity discussed below:
① When \ (n-\) is odd, then \ (n + 1 \) is an even number,
Seen from the title \ (n-A_ {+}. 1 = 2N--a_n. 1 \) , \ (A_ {n-2} + + + A_ {n-2N +. 1. 1} = \) ,
Two subtraction device, to give \ (A_ {n-2} + 2 = A_N + \) , i.e., odd-numbered columns to several other;
Therefore, before \ (60 \) entry in the sum of all the odd-numbered
\(S_{奇}=(a_1+a_3)+(a_5+a_7)+\cdots+(a_{57}+a_{59})=15\times 2=30\)；
② When \ (n-\) is even, then \ (n + 1 \) is an odd number,
Seen from the title \ (n-A_ {+}. 1. 1-2N = A_N + \) , then \ (A_ {n-2} + {-a_} = n-2N +. 1. 1 + \) ,
Adding the two equations to obtain \ (A_ {n-2} + 4N = A_N + \) , i.e. every two adjacent even-numbered arithmetic sum of columns;
Therefore, before \ (60 \) entry in the sum of all the even-numbered
\(S_{偶}=(a_2+a_4)+(a_6+a_8)+\cdots+(a_{58}+a_{60})\)
\(=4\times 2+4\times 6+4\times 10+\cdots+4\times 58\)
\(=4(2+6+10+\cdots+58)\)
\(=4\times\cfrac{(2+58)\times 15}{2}=1800\)；
Therefore \ (of S_ {60} = 1800 + 30 = 1830 \) .
Analysis: If the number of the column contains factor \ ((-. 1) n-^, (-. 1). 1-n-^ {} \) , and the general terms and associated summation method, as \ (S_n = (1-2) + (3-4) + (5-6) + \ cdots + (-. 1). 1-n-^ {} \ n-CDOT \) , for the additional \ (n-\) parity discussed.
Resolution:
When \ (\ n-) is an even number, \ (S_n = 1-2 + 3-4 + 5-6 + \ cdots + (-. 1). 1-n-^ {} \ n-CDOT \)
\(=(1-2)+(3-4)+(5-6)+\cdots+[(n-1)-n]\)，
\(=(-1)\times \cfrac{n}{2}\)；
When \ (n-\) is odd, \ (S_n = 1-2 + 3-4 + 5-6 + \ cdots + (-. 1). 1-n-^ {} \ n-CDOT \)
\(=(1-2)+(3-4)+(5-6)+\cdots+[(n-2)-(n-1)]+n\)，
\(=(-1)\times \cfrac{n-1}{2}+n=\cfrac{n+1}{2}\)；
Method 1: The conversion of packets summation method is as follows,
\(S=-(1^2+3^2+5^2+\cdots+99^2)+(2^2+4^2+6^2+\cdots+100^2)\)，
After such conversion, based on actual student learning, the basic idea of this stagnation;
Method 2: SUM and method, the following conversion
\(S=-1^2+2^2-3^2+4^2+\cdots-99^2+100^2\)
\(=100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)，
\(=(100-99)(100+99)+(98-97)(98+97)+\cdots+(2-1)(2+1)\)
\(=(100+99)+(98+97)+\cdots+(2+1)=5050\)
Analysis: available from known, when \ (n = 2 \) when, \ (+ A_2 A_3 = \ cfrac. 1} {2} {2 ^ \) , \ (n-=. 4 \) when, \ (+ A_4 = A_5 \ cfrac. 1 {{}}. 4 ^ 2 \) ,
\(S_{2n+3}=a_1+(a_2+a_3)+(a_4+a_5)+\cdots+(a_{2n}+a_{2n+1})+(a_{2n+2}+a_{2n+3})\)
\ (= 1+ \ cfrac {1} {2 ^ 2} + \ cfrac {1} {2 ^ 4} + \ cfrac {1} {2 ^ 6} + \ cdots + \ cfrac {1} {2 ^ {2n }} + \ cfrac {1} {2 ^ {2n + 2}} \)
\ (= (\ Cfrac {1} {2}) ^ 0 + (\ cfrac {1} {2}) ^ 2 + (\ cfrac {1} {2}) ^ 4 + (\ cfrac {1} {2 }) ^ 6+ \ cdots + (\ cfrac {1} {2}) ^ {2n} + (\ cfrac {1} {2}) ^ {2n + 2} \)
\(=\cfrac{1-[(\cfrac{1}{2})^2]^{n+2}}{1-(\cfrac{1}{2})^2}\)
\(=\cfrac{4}{3}(1-\cfrac{1}{4^{n+2}})\)；
Description: In this problem the method for finding the number of items, a number of \ (R & lt = \ cfrac {(2N + 2) + -0. 1} {2} = n-2 + \) .