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STEP I 1995 question 3 solution
From The Student RoomTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1995 question 3 solution
The final step in the reasoning comes from the elimination of terms due to alternating signs throughout.
Hence:
Hence:
Hence:
(The reason for the addition of "1" in the later stages of this solution was because the summation from 1 to n of "2n + 1" does not include the |
![\displaystyle \sum ^{n} _{r = 1} \{ f(r) - f(r - 1) \} \\ = \left[ f(1) - f(0) \right] + \left[ f(2) - f(1) \right] + \left[ f(3) - f(2) \right] + ... + \left[ f(n) - f(n - 1) \right] \\ = f(n) - f(0)](http://thestudentroom.co.uk/../latexrender/pictures/e24a6f2e72d19537f904cee930f1f6f9.png "\displaystyle \sum ^{n} _{r = 1} \{ f(r) - f(r - 1) \} \\ = \left[ f(1) - f(0) \right] + \left[ f(2) - f(1) \right] + \left[ f(3) - f(2) \right] + ... + \left[ f(n) - f(n - 1) \right] \\ = f(n) - f(0)")
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![f(r) = r^{2} (r + 1)^{2} \\ \implies f(r) - f(r - 1) = r^{2} (r + 1)^{2} - r^{2} (r - 1)^{2} \\ = r^{2} \left[ (r + 1)^{2} - (r - 1)^{2} \right] \\ = 4 r^{3}](http://thestudentroom.co.uk/../latexrender/pictures/bfefe21d2048857acad1cb458ec1203f.png "f(r) = r^{2} (r + 1)^{2} \\ \implies f(r) - f(r - 1) = r^{2} (r + 1)^{2} - r^{2} (r - 1)^{2} \\ = r^{2} \left[ (r + 1)^{2} - (r - 1)^{2} \right] \\ = 4 r^{3}")
![\displaystyle \sum _{r = 1} ^{n} r^{3} \\ = \frac{1}{4} \left[ f(n) - f(0) \right] \\ = \frac{1}{4} \left[ n^{2} (n + 1)^{2} \right] \\ = \frac{ n^{2} }{4} (n + 1)^{2}](http://thestudentroom.co.uk/../latexrender/pictures/18defe19b931f5e3d6326cb7038f342e.png "\displaystyle \sum _{r = 1} ^{n} r^{3} \\ = \frac{1}{4} \left[ f(n) - f(0) \right] \\ = \frac{1}{4} \left[ n^{2} (n + 1)^{2} \right] \\ = \frac{ n^{2} }{4} (n + 1)^{2}")
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![\displaystyle 1^{3} - 2^{3} + 3^{3} - 4^{3} + ... + (2n + 1)^{3} \\ = (-2n)^{3} + (-2n + 2)^{3} + ... + (-2)^{3} + 1^{3} + 3^{3} + ... + (2n + 1)^{3} \\ = 1 + \sum _{r = 1} ^{n} (2r + 1)^{3} - \sum _{r = 1} ^{n} (2r)^{3} \\ = 1 + \sum _{r = 1} ^{n} \left[ 8r^{3} + 12r^{2} + 6r + 1 \right] - \sum _{r = 1} ^{n} \left[ 8r^{3} \right] \\ = 1 + \sum _{r = 1} ^{n} \left[ 12r ^{2} + 6r + 1 \right]](http://thestudentroom.co.uk/../latexrender/pictures/0051d1ea694c3c2604655afef0daccf2.png "\displaystyle 1^{3} - 2^{3} + 3^{3} - 4^{3} + ... + (2n + 1)^{3} \\ = (-2n)^{3} + (-2n + 2)^{3} + ... + (-2)^{3} + 1^{3} + 3^{3} + ... + (2n + 1)^{3} \\ = 1 + \sum _{r = 1} ^{n} (2r + 1)^{3} - \sum _{r = 1} ^{n} (2r)^{3} \\ = 1 + \sum _{r = 1} ^{n} \left[ 8r^{3} + 12r^{2} + 6r + 1 \right] - \sum _{r = 1} ^{n} \left[ 8r^{3} \right] \\ = 1 + \sum _{r = 1} ^{n} \left[ 12r ^{2} + 6r + 1 \right]")
![\displaystyle 1^{3} - 2^{3} + 3^{3} - 4^{3} + ... + (2n + 1)^{3} \\ = 1 + \sum _{r = 1} ^{n} \left[ 12r ^{2} + 6r + 1 \right] \\ = 2n(n + 1)(2n + 1) + 3n(n + 1) + n + 1 \\ = (n +1)^{2} (4n + 1)](http://thestudentroom.co.uk/../latexrender/pictures/d18e27650257bb97ee04a8fcd0db1546.png "\displaystyle 1^{3} - 2^{3} + 3^{3} - 4^{3} + ... + (2n + 1)^{3} \\ = 1 + \sum _{r = 1} ^{n} \left[ 12r ^{2} + 6r + 1 \right] \\ = 2n(n + 1)(2n + 1) + 3n(n + 1) + n + 1 \\ = (n +1)^{2} (4n + 1)")
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