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aussy123

  • 3 years ago

Can someone help me finish this problem? Medal Rewarded!

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  1. aussy123
    • 3 years ago
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  2. aussy123
    • 3 years ago
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    In attachment

  3. Dr.Professor
    • 3 years ago
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    Hi... Do you know what type of problem is that

  4. aussy123
    • 3 years ago
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    Yea its an induction problem.

  5. Dr.Professor
    • 3 years ago
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    Great.. I have a request for you... pls write the question here

  6. aussy123
    • 3 years ago
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    Ok Ill try my best to put it on here correctly

  7. Dr.Professor
    • 3 years ago
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    Thanks.. for that

  8. aussy123
    • 3 years ago
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    Ok Prove the statement by mathematical induction. 3 + 5 + 7 + . . . + (2n + 1) = n(n + 2) 1. proposition is true when n = 1, since n(n + 2) = 1(1 + 2) =3 2. We will assume that the proposition is true for a constant k = n so, 3 + 5 + 7 + . . . + (2k + 1) = __________(k + __________) 3. Then, 3 + 5 + 7 + . . . + (2k + 1) + (_____k + _____) = k(k + 2) + (________k + _______)

  9. Dr.Professor
    • 3 years ago
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    so 1st one,

  10. Dr.Professor
    • 3 years ago
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    for the statement n= 1, the state ment reduces to\[1^2= \frac { 1\cdot 2\cdot 3 }{ 6 } \] and is obviously true. Assuming the statement is true for n = k: \[{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ 4 }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } \] , we will prove that the statement must be true for n = k + 1: \[{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ (k+1) }^{ 2 }=\frac { (k+1)(k+2)(2k+3) }{ 6 } \] The left-hand side of (2) can be written as \[{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 }+{ (k+1) }^{ 2 } \] In view of (1), this simplies to: \[{ (1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 })+{ (k+1) }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } +{ (k+1) }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)(2k+1)+6{ (k+1) }^{ 2 } }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)[k(2k+1)+6{ (k+1) }^{ 2 }] }{ 6 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)({ 2k }^{ 2 }+7k+6) }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k + 1)(k + 2)(2k + 3) }{ 6 } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \]

  11. Dr.Professor
    • 3 years ago
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    for the last part as its not clear

  12. Dr.Professor
    • 3 years ago
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    \[{ (1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 })+{ (k+1) }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } +{ (k+1) }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)(2k+1)+6{ (k+1) }^{ 2 } }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)[k(2k+1)+6{ (k+1) }^{ 2 }] }{ 6 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)({ 2k }^{ 2 }+7k+6) }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)(k+2)(2k+3) }{ 6 } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \]

  13. Dr.Professor
    • 3 years ago
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    Now lets solve the second one

  14. aussy123
    • 3 years ago
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    okay

  15. aussy123
    • 3 years ago
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    I realy dont know what you did here, but My work didnt ask for all this. Its really confusing.

  16. Dr.Professor
    • 3 years ago
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    The solution for 2

  17. Dr.Professor
    • 3 years ago
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    or goto this http://home.cc.umanitoba.ca/~thomas/Courses/textS1-21.pdf

  18. aussy123
    • 3 years ago
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    ok Ill try this..

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