STEP I 2003 question 1 solution - The Student Room
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STEP I 2003 question 1 solution

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 \displaystyle \sum _{r = -1} ^{n} r^{2} = pn^{3} + qn^{2} + rn + s

 \displaystyle n = -1 \implies \sum _{r = -1} ^{-1} r^{2} = -p + q - r + s = 1

 \displaystyle n = 0 \implies \sum _{r = -1} ^{0} r^{2} = s = 1

 \displaystyle n = 1 \implies \sum _{r = -1} ^{1} r^{2} = p + q + r + s = 2

 \displaystyle n = 2 \implies \sum _{r = -1} ^{2} r^{2} = 8p + 4q + 2r + s = 6

Hence:

 s + q - p - r = 1

 s = 1

 p + q + r + s = 2

 8p + 4q + 2r + s = 6

Hence:

 q - p - r = 0 = p - q + r , and:

 p + q + r = 1

Hence:

 2q = 1 \implies q = \frac{1}{2}

Hence:

 p + r = \frac{1}{2} .

So:

 8p + 8r = 4

 8p + 4q + 2r + s = 6 \implies 8p + 2r = 3

Hence:

 6r = 1 \implies r = \frac{1}{6} .

Hence:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
p = \frac{2]{6} = \frac{1}{3}

Hence:

 \displaystyle \sum _{r = -1} ^{n} r^{2} = \frac{1}{3} n^{3} + \frac{1}{2} n^{2} + \frac{1}{6} n + 1 .

As:

 \displaystyle \sum _{r = 0} ^{n} r^{2} = \sum _{r = -1} ^{n} r^{2} - (-1)^{2} = \sum _{r = -1} ^{n} r^{2} - 1

 \displaystyle \sum _{r = 0} ^{n} r^{2} = \frac{1}{3} n^{3} + \frac{1}{2} n^{2} + \frac{1}{6} n + 1 - 1 = \frac{1}{3} n^{3} + \frac{1}{2} n^{2} + \frac{1}{6} n

Hence:

 \displaystyle \sum _{r = 0} ^{n} r^{2} = \frac{1}{6} n \left( 2n^{2} + 3n + 2 \right) = \frac{1}{6} n \left( n + 1 \right) \left( 2n + 1 \right) .


 \displaystyle \sum _{r = -2} ^{n} r^{3} = an^{4} + bn^{3} + cn^{2} + dn + e

Hence:

 \displaystyle n = -2 \implies \sum _{r = -2} ^{-2} r^{3} = 16a - 8b + 4c - 2d + e = -8

 \displaystyle n = -1 \implies \sum _{r = -2} ^{-1} r^{3} = a - b + c - d + e = -9

 \displaystyle n = 0 \implies \sum _{r = -2} ^{0} r^{3} = e = -9

 \displaystyle n = 1 \implies \sum _{r = -2} ^{1} r^{3} = a + b + c + d + e = -8

 \displaystyle n = 2 \implies \sum _{r = -2} ^{2} r^{3} = 16a + 8b + 4c + 2d + e = 0

Hence:

 16a - 8b + 4c - 2d + e = -8

 a - b + c - d + e = -9

 e = -9

 a + b + c + d + e = -8

 16a + 8b + 4c + 2d + e = 0

Substituting "e" into the relevant equations produces:

 16a - 8b + 4c - 2d = 1

 a - b + c - d = 0

<latex? a + b + c + d = 1 </latex>

 16a + 8b + 4c + 2d = 9

Hence:

 16a - 16b + 16c - 16d = 0

Then:

 8b - 12c + 14d = 1 .

As:

 16a + 8b + 4c + 2d = 9 , and:

 16a + 16b + 16c + 16d = 16

 8b + 12c + 14d = 7 .

Coupling the above with the fact that  8b - 12c + 14d = 1 implies:

 24c = 6 \implies c = \frac{1}{4}

Hence:

 8b + 14d = 4 , and:

 a + b + d = \frac{3}{4} , and:

 16a + 8b + 2d = 8 .

Also:

 a - b - d = \frac{-1}{4}

Hence:

 2b + 2d =  1

Therefore:

 14b + 14d = 7

Then:

 6b = 3 \implies b = \frac{1}{2}

Hence:

 14d = 0 \implies d = 0

Hence:

 a = \frac{1}{4}

Hence:

 \displaystyle \sum _{r = -2} ^{n} r^{3} = \frac{1}{4} n^{4} + \frac{1}{2} n^{3} + \frac{1}{4} n^{2} - 9 .

As:

 \displaystyle \sum _{r = 0} ^{n} r^{3} = \sum _{r = -2} ^{n} r^{3} + 9

 \displaystyle \sum _{r = 0} ^{n} r^{3} = \frac{1}{4} n^{4} + \frac{1}{2} n^{3} + \frac{1}{4} n^{2} = \frac{1}{4}n^{2}(n^{2} + 2n + 1) = \frac{1}{4}n^{2}(n + 1)^{2} .

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