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

Clearly, the recursion relation for your sequence is
an+1 = an + n, with a1=1.

Since the n-th and (n+1)-th term differ by n, we suspect (from calculus) that the n-th term an has as its leading term n²/2. To ensure the leading term is divisible by 2, modify this as n(n-1)/2:
an=n(n-1)/2 + c for some unknown constant c.
After some trial and error, I found c=1, so

an=n(n-1)/2+1
or an=(n²-n+2)/2

Therefore, the 125-th term is
a125=125*124/2+1 = 7751

Similarly, you find the partial sum
∑an=∑(n²-n+2)/2
Again, calculus indicates that the leading term is of the form n³/6. After quite a bit of trial and error, I got

∑an=(n³+5n)/6
so that
n=1125 an=(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
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

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