Given sequence is 1,2,4,7,11,16... Find 125th term and also find sum.(good luck)

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# 6 Answers

Clearly, the recursion relation for your sequence is

a

_{n+1}= a_{n}+ n, with a_{1}=1.Since the n-th and (n+1)-th term differ by n, we suspect (from calculus) that the n-th term a

_{n}has as its leading term n²/2. To ensure the leading term is divisible by 2, modify this as n(n-1)/2:a

_{n}=n(n-1)/2 + c for some unknown constant c.After some trial and error, I found c=1, so

a

_{n}=n(n-1)/2+1or a

_{n}=(n²-n+2)/2Therefore, the 125-th term is

a

_{125}=125*124/2+1 = 7751Similarly, you find the partial sum

∑a

_{n}=∑(n²-n+2)/2Again, calculus indicates that the leading term is of the form n³/6. After quite a bit of trial and error, I got

∑a

_{n}=(n³+5n)/6so that

∑

_{n=1}^{125}a_{n}=(125³+5(125))/6 = 325,625.1,2,4,7,11,16...

1, 2, 3, 4, 5,..., first layer difference

Adding the difference to the original sequence, you get the next term.

Therefore, the nth term

t(n)

= 1+(0+1+2+3+...+(n-1))

= 1+(n/2)(n-1)

= (1/2)(n^2-n+2)

Sum from 1 to n

= S(n)

= ∑(i = 1, n) (1/2)(i^2-i+2)

= (1/2)[(1/6)n(n+1)(2n+1) - (1/2)n(n+1) + 2n]

Check:

t(1) = 1, S(1) = 1

t(2) = 2, S(2) = 3

...

t(125) = 7751, S(125) = 325625 <==Answer

Hey Prasenjeet -- a "back-of-the-napkin" approach ...

1,2,3,4,5 => 1,2,4,7,11 ... t(5) = sum of (1=>5) minus 4 ...

t(n) = mid value * n minus (n-1) = 1/2(n+1)n less (n-1) ...

t(125) = 63*125 less 124 = 7500 plus 375 less 124 =

**7751**...t(n) ~ n*n/2 less n ==> for sum, Integrate [n*n/2 less n] ...

sum ~ 1/6(n^3) less 1/2(n^2) ~ 1/6(125) cubed ...

sum ~ 1/6(120)130(125) ~ 2500(130) ~ 10k(130)/4 ~ 1300k/4 ~

**325k**:)here is a different approach to finding the 125th term

1,2,4,7,11,16,22,29,37,... is the sequence

the difference between the 8th and 9th terms is 8

the difference between the 7th and 8th term is 7

therefore the difference between the 124th term and the 125th term is 124

if we add the differences: 1+2=3, 1 less than the 3rd term

1+2+3=6, 1 less than the 4th term

1+2+3+4=10, 1 less than the 5th term

the difference between the 125th term and the 124th term is 124 by looking at the other differences

the sum of all the differences will be 1 less than the 125th term

so we add all the differences and add 1 to get the 125th term

so we add 1+2+3+4+5+6+...+119+120+121+122+123+124

add 1+124=125, 2+123=125, 3+122=125 and so on

how many groups of 125 do we have ? we have 62 groups

therefore 125*62=7750 and now we add 1 to 7750 to get 7751, the 125th term

I used an excel spreadsheet and got the same answers as Robert.

1

1

1

2

2

3

4

4

7

5

11

6

16

7

22

8

29

9

37

10

46

11

56

12

67

13

79

14

92

15

106

16

121

17

137

18

154

19

172

20

191

21

211

22

232

23

254

24

277

25

301

26

326

27

352

28

379

29

407

30

436

31

466

32

497

33

529

34

562

35

596

36

631

37

667

38

704

39

742

40

781

41

821

42

862

43

904

44

947

45

991

46

1036

47

1082

48

1129

49

1177

50

1226

51

1276

52

1327

53

1379

54

1432

55

1486

56

1541

57

1597

58

1654

59

1712

60

1771

61

1831

62

1892

63

1954

64

2017

65

2081

66

2146

67

2212

68

2279

69

2347

70

2416

71

2486

72

2557

73

2629

74

2702

75

2776

76

2851

77

2927

78

3004

79

3082

80

3161

81

3241

82

3322

83

3404

84

3487

85

3571

86

3656

87

3742

88

3829

89

3917

90

4006

91

4096

92

4187

93

4279

94

4372

95

4466

96

4561

97

4657

98

4754

99

4852

100

4951

101

5051

102

5152

103

5254

104

5357

105

5461

106

5566

107

5672

108

5779

109

5887

110

5996

111

6106

112

6217

113

6329

114

6442

115

6556

116

6671

117

6787

118

6904

119

7022

120

7141

121

7261

122

7382

123

7504

124

7627

125

7751

325625 Total

Last Number 7876 and sum 333501. I wrote a visual basic program in Microsoft Excel.

Sub Macro1()

' Macro1 Macro

' Macro recorded 10/26/2013 by Owner'

Dim I As Integer

Dim t As Integer

Dim Next_number As Integer

Dim Last_number As Integer

Dim Sum As Double

Sum = 1

Last_number = 1

For I = 0 To 124

t = t + 1

Next_number = Last_number + t

Last_number = Next_number

Sum = Sum + Next_number

Next I

End Sub

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