IQ Tests in Logical Thinking : Number Sequences II
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SEQUENCE  EXPLANATION  1, 3, 6, 10, 15, 21, 28, 36  1; 1 + 2 = 3; 1 + 2 + 3 = 6; 1 + 2 + 3 + 4 = 10; 1 + 2 + 3 + 4 + 5 = 15; 1 + 2 + 3 + 4 + 5 + 6 = 21; 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28; 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36; The nth term in the sequence is given by n (n + 1)/2, and the numbers are often referred to as triangular numbers.  1, 3, 7, 15, 31, 63, 127, 255  2^{1}  1 = 2  1 = 1; 2^{2}  1 = 4  1 = 3; 2^{3}  1 = 8  1 = 7; 2^{4}  1 = 16  1 = 15; 2^{5}  1 = 32  1 = 31; 2^{6}  1 = 64  1 = 63; 2^{7}  1 = 128  1 = 127; 2^{8}  1 = 256  1 = 255; The nth term in the sequence is given by 2^{n}  1. So, the difference between two consecutive numbers forms the following simple sequence: 2, 4, 8, 16, 32, 64, 128, ...  0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55  0 + 1 = 1; 1 + 1 = 2; 1 + 2 = 3; 2 + 3 = 5; 3 + 5 = 8; 5 + 8 = 13; 8 + 13 = 21; 13 + 21 = 34; 21 + 34 = 55; Each term (starting with the third term) in the sequence is the sum of the two terms preceding it. The series is often referred to as the Fibonacci series. Fibonacci (1175) believed that this series was followed by various natural phenomena. In fact, the number of leaves on the stems of particular plants follows this series.  15, 12, 24, 20, 33, 28, 42, 36  The odd terms of the sequence continually increase by 9, i.e., 15 + 9 = 24; 24 + 9 = 33 ; 33 + 9 = 42; ... The even terms of the sequence continually increase by 8, i.e., 12 + 8 = 20; 20 + 8 = 28; 28 + 8 = 36; ...  2, 6, 12, 20, 30, 42, 56, 72  (1)(2) = 2; (2)(3) = 6; (3)(4) = 12; (4)(5) = 20; (5)(6) = 30; (6)(7) = 42; (7)(8) = 56; (8)(9) = 72; The nth term in the sequence is given by n (n + 1).  6, 24, 60, 120, 210, 336, 504, 720  (1)(2)(3) = 6; (2)(3)(4) = 24; (3)(4)(5) = 60; (4)(5)(6) = 120; (5)(6)(7) = 210; (6)(7)(8) = 336; (7)(8)(9) = 504; (8)(9)(10) = 720; The nth term in the sequence is given by n (n + 1) (n + 2).  1, 2, 6, 24, 120, 720  1 = 1; (1)(2) = 2; (1)(2)(3) = 6; (1)(2)(3)(4) = 24; (1)(2)(3)(4)(5) = 120; (1)(2)(3)(4)(5)(6) = 720; The nth term in the sequence is given by n! (factorial of n), which is defined as the product of all integers from 1 to n.  0, 1, 2, 7, 20, 61, 182, 547  3(0) + 2(1) = 2; 3(1) + 2(2) = 7; 3(2) + 2(7) = 20; 3(7) + 2(20) = 61; 3(20) + 2(61) = 182; 3(61) + 2(182) = 547; The nth term in the sequence is given by t_{n} = 3 t_{n  2} + 2 t_{n  1}. A term (starting with the third) in the sequence is a linear combination of the preceding two terms. So, let the nth term in the sequence be given by t_{n} = a t_{n  2} + b t_{n  1}. For n = 3, 2 = a(0) + b(1) For n = 4, 7 = a(1) + b(2) Thus, a = 3 and b = 2.  1/4, 0, 1, 3, 13, 51, 205, 819  4(1/4)  3(0) = 1; 4(0)  3(1) = 3; 4(1)  3(3) = 13; 4(3)  3(13) = 51; 4(13)  3(51) = 205; 4(51)  3(205) =819; The nth term in the sequence is given by t_{n} = 4 t_{n  2}  3 t_{n  1}. A term (starting with the third) in the sequence is a linear combination of the preceding two terms. So, let the nth term in the sequence be given by t_{n} = a t_{n  2} + b t_{n  1}. For n = 3, 1 = a(1/4) + b(0) For n = 4, 3 = a(0) + b(1) Thus, a = 4 and b = 3.  1, 10, 11, 100, 101, 110, 111, 1000  1 (base 2) = 1(1) = 1 (base 10); 10 (base 2) = 1(2) + 0(1) = 2 (base 10); 11 (base 2) = 1(2) + 1(1) = 3 (base 10); 100 (base 2) = 1(4) + 0(2) + 0(1) = 4 (base 10); 101 (base 2) = 1(4) + 0(2) + 1(1) = 5 (base 10); 110 (base 2) = 1(4) + 1(2) + 0(1) = 6 (base 10); 111 (base 2) = 1(4) + 1(2) + 1(1) = 7 (base 10); 1000 (base 2) = 1(8) + 0(4) + 0(2) + 0(1) = 8 (base 10); The sequence is simply 1, 2, 3, 4, 5, ... in the binary (base 2) system.  1, 2, 10, 37, 101, 226  2  1 = 1; 10  2 = 8; 37  10 = 27; 101  37 = 64; The differences between two consecutive numbers are 1, 8, 27, 64, ... (cubes of integers starting with 1). So, 101 + 5^{3} = 101 + 125 = 226  2, 5, 10, 17, 26, 37, 50, 65  The terms are merely one more than the squares of integers starting with 1. Thus, 1^{2} + 1 = 1 + 1 = 2; 2^{2} + 1 = 4 + 1 = 5; 3^{2} + 1 = 9 + 1 = 10; 4^{2} + 1 = 16 + 1 = 17; 5^{2} + 1 = 25 + 1 = 26; 6^{2} + 1 = 36 + 1 = 37; Alternatively, the differences between consecutive terms form the following simple sequence: 3, 5, 7, 9, 11, 13, 15, ...  7, 26, 63, 124, 215, 342  The terms are merely one less than the cubes of integers starting with 2. Thus, 2^{3}  1 = 8  1 = 7; 3^{3}  1 = 27  1 = 26; 4^{3}  1 = 64  1 = 63; 5^{3}  1 = 125  1 = 124; 6^{3}  1 = 216  1 = 215; 7^{3}  1 = 343  1 = 342;  2, 12, 36, 80, 150, 252, 392, 576  The terms are the sum of the squares and cubes of integers, starting with 1. Thus, 1^{2} + 1^{3} = 1 + 1 = 2; 2^{2} + 2^{3} = 4 + 8 = 12; 3^{2} + 3^{3} = 9 + 27 = 36; 4^{2} + 4^{3} = 16 + 64 = 80; 5^{2} + 5^{3} = 25 + 125 = 150; 6^{2} + 6^{3} = 36 + 216 = 252; 7^{2} + 7^{3} = 49 + 343 = 392; 8^{2} + 8^{3} = 64 + 512 = 576;  2, 5, 17, 65, 257, 1025  The terms are merely one more than the powers of 4. Thus, 4^{0} + 1 = 1 + 1 = 2; 4^{1} + 1 = 4 + 1 = 5; 4^{2} + 1 = 16 + 1 = 17; 4^{3} + 1 = 64 + 1 = 65; 4^{4} + 1 = 256 + 1 = 257; 4^{5} + 1= 1024 + 1 = 1025;  9, 729, 8, 512, 7, 343  The odd terms are merely the integers starting with 9 in descending order. The even terms are the cubes of the odd terms. Thus, 9^{3} = 9 x 9 x 9 = 729; 8^{3} = 8 x 8 x 8 = 512; 7^{3} = 7 x 7 x 7 = 343;  361, 289, 225, 169, 121  The terms are merely the squares of odd integers starting with 19 in descending order. Thus, 19^{2} = 19 x 19 = 361; 17^{2} = 17 x 17 = 289; 15^{2} = 15 x 15 = 225; 13^{2} = 13 x 13 = 169; 11^{2} = 11 x 11 = 121; The differences between two consecutive numbers are 72, 64, 56, 48, ... (a simple sequence starting with 72 and continually decreasing by 8).  1/12, 1/2, 9/8, 13/6, 17/4, 21/2  The numerators (starting with 1) increase continually by 4, and the denominators (starting with 12) decrease continually by 2. The numerators are 1, 5, 9, 13, 17, 21. The denominators are 12, 10, 8, 6, 4, 2. So, the fractions are 1/12, 5/10, 9/8, 13/6, 17/4, and 21/2. Note that 5/10 is equivalent to 1/2. 
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