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Homework Statement
This is taken from STEP I 1990, Q4.
(i) The sequence a1, a2, ..., an, ... forms an arithmetic progression. Establish a formula, involving n, a1, and a2, for the sum of the first n terms.
(ii) A sequence b1, b2, ..., bn, ... is called a double arithmetic progression if the sequence of differences, b2 - b1, b3 - b2, ..., bn+1 - bn, ... is an arithmetic progression. Establish a formula, involving n, b1, b2 and b3, for the sum b1 + b2 + ... + bn of the first n terms of such a progression.
(iii) A sequence c1, c2, ..., cn, ... is called a factorial progression if cn+1 - cn = n!d, for some non-zero d and every n ≥ 1. Suppose 1, b2, b3, ... is a double arithmetic progression, and also that b2, b4, b6 and 220 are the first four terms in a factorial progression. Find the sum 1 + b1 + b2 + ... + bn.
Homework Equations
Standard arithmetic progression formulae below
The nth term of an AP: un = a + (n-1)d
The sum of the first n terms of an AP: Sn = (n/2)(a + l) = (n/2)(2a + (n-1)d)
The Attempt at a Solution
I've done (i) quite comfortably and got
[tex]\frac{n}{2}((3-n)a_{1} + (n-1)a_{2})[/tex]
However, (ii) is where I get stuck. By considering the sequence of differences, I've established that
[tex]b_n = a + (n-2)d + b_{n-1}[/tex]
with a = b2 - b1, and d = (b3 - b2) - (b2 - b1). Can anyone guide me on where to go next?
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