STEP I 1990 question 4 solution

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Arithmetic Progressions

If, $\displaystyle a_{1}, a_{2}, \; ... \;, a_{n}$ is an arithmetic progression, then

$\displaystyle S_{n} = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(2a_{1} + (n-1)(a_{2}-a_{1})) = \frac{n}{2}((3-n)a_{1} + (n-1)a_{2})$

Double Arithmetic Progressions

If $\displaystyle b_{2}-b_{1}, \; b_{3}-b_{2}, \; ... \;, b_{n+1}-b_{n}, \; ...$ is an arithmetic progression, then

$\displaystyle b_{n} = a + (n-2)d + b_{n-1}$

$\displaystyle \Rightarrow b_{n} = \left( \sum^{n}_{r=2} a + (r-2)d\right) + b_{1} = (n-1)a + d \left( \sum^{n}_{r=2}r-2 \right) + b_{1}$

$\displaystyle \sum^{n}_{r=2}(r-2) = \sum^{n-2}_{k=1} k = \frac{1}{2}(n-1)(n-2)$

$\displaystyle \Rightarrow b_{n} = (n-1)a + \frac{1}{2}(n-1)(n-2)d + b_{1}$

$\displaystyle \sum_{r=1}^{n} b_{r} = a\left(\sum_{r=1}^{n} n-1\right) + \frac{d}{2}\left(\sum_{r=1}^{n} n^{2} - 3n + 2 \right) + nb_{1}$

$\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{2}n(n-1)a + \frac{1}{6}n(n-1)(n-2)d + nb_{1}$

$\displaystyle a = b_{2} - b_{1}, \; d = (b_{3} - b_{2}) - (b_{2} - b_{1}) = b_{3} - 2b_{2} + b_{1}$

Several minutes of tedious algebra later:

$\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{6}n\left((n^{2} - 6n + 11)b_{1} - (n-1)(2n-7)b_{2} + (n-1)(n-2)b_{3}\right)$

Factorial Progressions

$b_{4}-b_{2} = d, \; b_{6}-b_{4} = 2d, \; 220 - b_{6} = 6d$

$b_{2} - 1 = a, \; b_{n} - b_{n-1} = a + (n-2)x$

$(b_{3} - b_{2}) + (b_{4} - b_{3}) = b_{4} - b_{2} = 2a + 3x = d$

$(b_{5}-b_{4}) + (b_{6} - b_{5}) = b_{6} - b_{4} = 2a + 7x = 2d$

$2a + 7x = 4a + 6x \Rightarrow x = 2a, \; d = 8a$

$(b_{2}-1) + (b_{3}-b_{2}) + \cdots + (b_{6}-b_{5}) = b_{6} - 1 = 5a + 10x$

$b_{6} = 220 - 6d = 5a + 10x + 1 \Rightarrow 5a + 10x + 6d = 219$

$5a + 20a + 48a = 219 \Rightarrow 73a = 219$

$\Rightarrow a = 3, \; x = 6, \; d = 24$

Putting $a = 3, \; d = 6, \; b_{1} = 1$ into

$\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{2}n(n-1)a + \frac{1}{6}n(n-1)(n-2)d + nb_{1}$

we get:

$\displaystyle S_{n} = \frac{3}{2}n(n-1) + n(n-1)(n-2) + n = \frac{1}{2}n(2n^{2} - 3n + 3)$

Solution by Dystopia.

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