Compound Arithmetic Sequence
a0 a1 a2 a3 a4 a5 a6
b0 b1 b2 b3 b4 b5 ------- 1st differences
c0 c1 c2 c3 c4 -------------- 2nd differences
d d d d ------------------- 3rd differences = constant
We have here an irregular sequence (a_n) = (a_1, a_2, a_3, ..., a_k, ...).
The differences between the elements of (a_n), the 1st differences, are non-constant.
So also are the differences between the elements of (b_n), the 2nd differences.
However, the differences between the elements of (c_n) are constant. This makes (c_n) a regular arithmetic sequence, and so we can write,
cn = c0 + nd, n = 0,1,2,3,...
The sequence (b_n) is non-regular, but we can write,
b_(n+1) = b_n + c_n
Using the expression for the c-sequence,
b_(n+1) = b_n + c_0 + nd
Solve the recurrence relation for b_n
Develop the terms of the sequence.
b1 = b0 + c0
b2 = b1 + c0 + 1.d = b0 + 2.c0 + 1.d
b3 = b2 + c0 + 2.d = b0 + 3.c0 + (1 + 2).d
b4 = b3 + c0 + 3.d = b0 + 4.c0 + (1 + 2 + 3).d
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b(n+1) = b_0 + (n+1).c0 + sum[k=1..n](k) * d
b_(n+1) = b_0 + (n+1).c0 + (nd/2)(n + 1)
b_(n+1) = b_0 + (n+1)(c_0 + nd/2)
We could also write c0 = b1 – b0 in the above expression.
Now that we have a rule/expression for the general term of the b-sequence, let us analyse the a-sequence.
Since the sequence (a_n) is non-regular like (b_n), we can write,
a_(n+1) = a_n + b_n
Substituting for the general term of the b-sequence,
a_(n+1) = a_n + b0 + n(c0 + (n-1)d/2) (co can be replaced later with c0 = b1 – b0)
Solve the recurrence relation for a_n
Develop the terms of the sequence.
a1 = a0 + b0
a2 = a1 + b0 + 1.c0 + 1.0.d/2 = a0 + 2.b0 + 1.c0 + 1.0.d/2
a3 = a2 + b0 + 2.c0 + 2.1.d/2 = a0 + 3.b0 + (1 + 2).c0 + (1.0 + 2.1).d/2 (N.B. here, 1.0 = 1*0, 2.1 = 2*1)
a4 = a3 + b0 + 3.c0 + 3.2.d/2 = a0 + 4.b0 + (1 + 2 + 3).c0 + (1.0 + 2.1 + 3.2).d/2
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a_n = a0 + n.b0 + sum[k=1..n-1] k * c0 + sum[k=1..n-1](k(k-1)) * (d/2)
a_n = a0 + n.b0 + (c0/2).n(n-1) + sum[k=1..n-1](k^2) * (d/2) – sum[k=1..n-1](k) * (d/2)
a_n = a0 + n.b0 + (c0.n/2)(n-1) + (d/2){(n/6)(n-1)(2n-1) – (n/2)(n-1)}
a_n = a0 + n.b0 + (c0.n/2)(n-1) + (nd/6)(n^2 – 3n + 2)
Substituting for c0 = b1 – b0, b1 = a2 – a1, b0 = a1 – a0
Then, c0 = a0 – 2a1 + a2.
These substitutions give,
a_n = (1 + (n/2)(n – 3)).a0 – n(n – 2).a1 – (n/2)(1 – n).a2 + (nd/6)(n^2 – 3n + 2)
Now substituting for a0 = 2, a1 = 5, a2 = 9 and d = 5, the rule is,
a_n = 2 + n(n – 3) – 5n(n – 2) – (9/2).n.(1 – n) + (5/6).n.(n^2 – 3n + 2)