STEP I 2003 question 1 solution

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STEP I 2003 question 1 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2003 question 1 solution

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

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

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

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

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

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 6 ]

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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