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Carto Wong Notes for HKCEE Mathematics Arithmetic and Geometric Sequence/Series Recall that a sequence is a function defined on {1, 2, 3, } and usually denoted by one of the following notations: {xn }n 1 , {xn } , {x1 , x2 , x3 , } . For example, {2n 1}n 1 , {2n 1} and {1, 3, 5, } represent the same sequence. Definition. (AS and GS) A sequence {xn } is called an arithmetic sequence（算術數列） if ( n 1, 2, 3, ), xn1 xn d where d is a constant. Arithmetic sequence is abbreviated to “A.S.”. A sequence {xn } is called a geometric sequence（幾何數列） if ( n 1, 2, 3, ), xn1 rxn where r is a constant. Geometric sequence is abbreviated to “G.S.”. d is called common difference（公差） and r is called common ratio（公比）. General Term and Partial Sum. The general term of an A.S. {xn } with common difference d is xn x1 (n 1)d . The nth partial sum is x1 xn n( x1 xn ) n [2 x1 (n 1)d ] . 2 2 The general term of a G.S. { yn } with common ratio r is yn y1 r n 1 . The nth partial sum is y1 yn y1 (1 r n ) . 1 r If 1 r 1 , the sum to infinity is y1 y2 y1 . 1 r Page 1 of 6 Carto Wong Basic Techniques and Typical Examples Example 1. Let {xn } be an A.S. with x4 3 and common difference 5. Find the general term. Solution. The general term is xn x1 5(n 1) . Put n 4 , we have 3 x1 5 3 x1 12 . Hence, the general term is xn 12 5(n 1) 5n 17 . The general term of any A.S. is of the form xn pn q (p, q are constants). TIP. Moreover, the constant p should be the common difference. If the answer you obtained is not of this form, then it must be wrong. Example 2. Let {xn } be an A.S. with the nth partial sum S n n 2 3n . Find the general term. Solution. The general term is xn Sn Sn 1 [(n 1) 2 3(n 1)] [n 2 3n] 2n 2 . TIP. The nth partial sum S n of an A.S. must be a quadratic expression of n with constant term zero. This allows us to adopt the convention S0 0 . By examples 1 and 2, we know that the following information of an A.S. are equivalent: (in the sense that if we know one then we know the others) (i.) Particular term and common difference; (ii.) Partial sum; (iii.) General term. Page 2 of 6 Carto Wong Example 3. Let { yn } be a G.S. with y4 2 and common ratio 3. Find the general term. Solution. The general term is yn 3n 1 y1 . Put n 4 , we have 2 33 y1 y1 Hence, the general term is yn 2 . 27 2 3n 1 . 27 Example 4. Let { yn } be a G.S. with the nth partial sum Tn 3n 1 . Find the general term. 2 Solution. The general term is yn Tn Tn 1 TIP. 3n 1 3n 1 1 3n 3n 1 2 3n 1 3n 1 . 2 2 2 2 Again, we adopt the convention T0 0 here. In the examination, the question may ask you to find the nth partial sum S n (of an A.S. or a G.S.). Once you get your answer, it is a good habit to put n 0 and check if S0 0 . For example, if your answer is Sn n 2 n 2 (note that S0 2 0 ) then it must be wrong. A similar figure in which we saw before: Page 3 of 6 Carto Wong Interesting Example. Let ABC be an equilateral triangle with area 1 cm 2 and A , B , C are respectively the midpoints of BC, CA, AB. The midpoints are joined to form ABC as shown in the figure. A C' B' (a) Find the area of ABC . B In each of the remaining triangles ACB , CBA , BAC . We draw their medial triangles (which are shaded in the figure). A (b) Find the area of each shaded smaller triangle. C' (c) Repeat the process indefinitely. Find the total area of shaded triangles. C A' B B' C A' Solution. (a) Since ABC is similar to ABC , AB AB C ABC AB 1 ABC 4 1 cm 2 4 2 (b) By similarity again, the area of each shaded smaller triangle is 1 3 9 (c) Total area is 4 16 64 1 4 3 1 4 1 1 1 cm 2 . 4 4 16 1 cm 2 . The figure in the right is called Sierpinski Triangle（西 爾平斯基三角）. It is an example of fractal（分形）in modern mathematics. The most interesting property of Sierpinski Triangle is that it has non-integral dimension log3 log 2 1.585 . Page 4 of 6 Carto Wong Additional Example. (Out of Syllabus) A sequence {xn } is defined recursively by x1 1 xn1 2 xn 1 (n 1, 2, ) Find the general term of this sequence. We write down the first few terms: n xn 1 2 3 4 5 6 7 1 3 7 15 31 63 127 This is neither A.S. nor G.S.! Solution. Let yn xn 1 , then y1 2 and the recurrence relation becomes yn 1 xn 1 1 (2 xn 1) 1 2( xn 1) 2 yn Note that { yn } is a G.S.! Its general term is yn 2n 1 y1 2n . Hence, xn 2n 1 is the general term of the original sequence. Further Investigation. More generally, suppose a sequence is defined by x1 a xn1 pxn q (n 1, 2, ) where a, p, q are given constants. (Note that A.S. and G.S. are sequences of this type.) It is natural to ask: how to find the general term of this sequence? (Think yourself and then see the solution in next page) Page 5 of 6 Carto Wong Solution. If p 1 then the sequence is an A.S., this case is easy. We now consider the case p 1 . q Let yn xn . Then { yn } is a G.S. with common ratio p. This is because p 1 q yn 1 xn 1 p 1 q ( pxn q ) p 1 q q p yn q p 1 p 1 pyn Hence, yn p n 1 y1 xn p n 1 y1 q q q p n 1 a . p 1 p 1 p 1 Exercise. Find the general term of the sequence {xn } defined by x1 2 xn1 3xn 5 (n 1, 2, ) The first few terms are: n xn Answer: 1 2 3 4 5 6 7 2 1 2 11 38 119 362 5 3n1 . 2 2 Appendix: why 0.9 = 1 ? What is the value of 0.999 ? May be somebody told you before the answer is 1. This surprising result can be proved as follows: The number 0.9 is, by definition, the sum of the series 0.9 0.09 0.009 This is a geometric series with common ratio 0.1, the sum is 0.9 1. 1 0.1 This example shows that the decimal representation（十進制表達式）of a number is not unique in general. Page 6 of 6