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What Is nth term of the sequence 4,7,10,16,28 ?

This is a hard nth term question! We will solve with method of differences. 4,7,10,16,28 First difference (relates to a linear sequence if constant n): 3,3,6,12 Second difference (relates to a quadratic if constant n^2): 0,3,6 Third difference (relates to a cubic if constant n^3): 3,3 The third difference is constant (though we only have two terms of it, if this wasnt constant, we couldn't easily solve nth term for this part of the sequence). Since third difference is constant, we know it is cubic of form an^3+bn^2+cn+d = nth term where a,b,c,d are constants. We can also do this with 5 simultaneous equations from here on out. first term when n = 1: a + b + c + d = 4 [1] second term when n = 2: 8a + 4b + 2c + d = 7 [2] third term n = 3: 27a + 9b + 3c + d = 10 [3] fourth term n = 4: 64a + 16b + 4c + d = 16 [4] fifth term n = 5: 125a + 25b + 5c + d = 28 [5] I have labelled all equations with square brackets (not part of the actual equation). Using [2] - [1]: 7a + 3b + c = 3 [2-1] Using [3] - [2]: 19a + 5b + c = 3 [3-2] Using [2-1] - [3-2]: 12a + 2b = 0 [A] Relabelled as letters Using [4] - [3]: 37a + 7b + c = 6 [4-3] Using [5] - [4]: 61a + 9b + c = 12 [5-4] Using [5-4] - [4-3]: 24a + 2b = 6 [B] Relabelled as letters [B] - [A]: 12a = 6 therefore a = 1/2 If a = 1/2 using [B]: 12+2b=6 therefore b = -6/2 = -3 Then using [2-1]: 7a + 3b + c = 3 which substituting a and b in goes to 7/2 + (3x-3) + c = 3 therefore c = 3 + 9 - 7/2 when you rearrange and expand c = 17/2 Finally lets use [1] to find d. a + b + c + d = 4 1/2 - 3 + 17/2 + d = 4 d = 4 - 18/2 + 3 d = -2 You should be able to get to this result using different combinations of these equations. Just make sure you use as many as you can. If you try and cancel to many terms using the same information (e.g same equations). You'll likely just end up with something like 0 = 0. Therefore nth term is combination of a,b,c,d as we defined earlier. an^3 + bn^2 + cn + d = 1/2(n^3) - 3(n^2) + (17/2)n - 2 Sorry for such a long solution. I think there are other ways to get to this solution. It is quite a tricky one and easy to make mistakes in the workings. You can test this solution and I think it works

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