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STEP I 1995 question 3 solutionTSR 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:
Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 1685x55
\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] Hence: Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 1029x55 .
\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) (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 |
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![\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://www.thestudentroom.co.uk/latexrender/pictures/36/36029787bff3e6cc0f1a86f545d4ec8a.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)")
![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://www.thestudentroom.co.uk/latexrender/pictures/12/127a10728146117fced4eef7e7d17ce8.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://www.thestudentroom.co.uk/latexrender/pictures/57/57b45414c8ae078e95df8e1418e15305.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}")
