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unicwaan

  • one year ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

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  1. unicwaan
    • one year ago
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    \[1^2+4^2+7^2+...+(3n−2)^2=\frac{ n(6n2−3n−1) }{ 2 }\]

  2. mathstudent55
    • one year ago
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    An induction proof works like this: 1. Show the formula works for n = 1 2. Assume it works for n = k 3. Show it works for n = k + 1

  3. mathstudent55
    • one year ago
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    Let's start. Can you show the formula works for n = 1? Copy the formula. Then rewrite the formula replacing n with 1. What do you get?

  4. mathstudent55
    • one year ago
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    \(\Large 1^2+4^2+7^2+...+(3n−2)^2=\dfrac{ n(6n2−3n−1) }{ 2 }\) Let n = 1 \(\Large [3(1)−2)]^2=\dfrac{ 1(6 \cdot 1^22−3 \cdot 1−1) }{ 2 }\) \(\Large 1 = \dfrac{1(2)}{2} \) \(\Large 1 = 1\) We proved the expression works for n = 1.

  5. mathstudent55
    • one year ago
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    Now we assume the expression is true for n = k: \(\Large 1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \dfrac{k(6k^2 - 3k - 1)}{2} \)

  6. mathstudent55
    • one year ago
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    Now we need o prove it works for n = k + 1 \(\Large 1^2 + 4^2 + 7^2 + ...+ (3k - 2)^2 + [3(k + 1) - 2]^2 =\) \(\Large = \dfrac{(k + 1) [6(k + 1)^2 - 3(k + 1) - 1]}{2} \) \(\Large = \dfrac{(k + 1)[6(k^2 + 2k + 1) - 3k -3 - 1]}{2} \) \(\Large = \dfrac{(k + 1)(6k^2 + 12k + 6 - 3k - 4)}{2} \) \(\Large = \dfrac{(k + 1)[6(k^2 + 2k + 1) - 3k - 3 - 1)}{2} \) \(\Large = \dfrac{(k + 1)[6(k + 1)^2 - 3(k + 1) - 1)}{2} \) The last expression above is the formula with k replaced by k + 1. This shows the formula works for n = k + 1. By induction, the formula has been proved to be true.

  7. madhu.mukherjee.946
    • one year ago
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    @mathstudent55 superb

  8. mathstudent55
    • one year ago
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    @madhu.mukherjee.946 Thanks!

  9. unicwaan
    • one year ago
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    Thank you for showing me the walk-through steps! Sorry that I went away from my compute! @mathstudent55

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