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    i am having mania with induction problems. i don't mind if that mania leads me to the correct answers but alas. i have attached a pdf file that conatins five of my difficulties. your help would be really appreciated. thanks.
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  1. File Type: pdf proof.pdf (25.2 KB, 96 views)
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    Here we go:

    sin x = x - x³/3! + x^5/5! - ...

    [x³/6 - x^5/120 + ...] = x - sin x

    but for x>0, [x³/6 - x^5/120 + ...] > 0

    therefore x - sin x > 0

    x > sin x (1)

    similarly

    sin x = x - x³/6 + x^5/120 - x^7/5040 + ...

    [x^5/120 - x^7/5040 + ...] = sin x - x + x³/6

    for x > 0 , [x^5/120 - x^7/5040 + ...] > 0

    therefore sin x - x + x³/6 > 0

    sin x > x - x³/6 (2)

    combine (1) and (2) gives

    x > sin x > x - x³/6

    as required
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    let the general (nth) term in a sequence be n³ + 6n² + 8n , denoted as Un
    then Un+1 = (n+1)³ + 6(n+1)² + 8(n+1)
    = n³ + 9n² + 23n + 15

    subtracting Un from this:

    Un+1 - Un = 3n² +15n +15
    = 3[n² + 5n + 5]

    therefore, the difference of two adjacent terms is always divisible by 3

    therefore, if Ui is divisible by 3, then so is Ui+1 (1)

    now U1 = 1 + 6 + 8 = 15 = 5x3

    therefore U1 is divisible by 3, and by line (1), so is U2 , U3 etc

    therefore n³ + 6n² + 8n is divisible by 3 for all positive integer values of n.
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    similar method as before, Un = 7^(2n) + [2^(3n -3)][3^(n-1)]

    =7^(2n) + [{2^(3n)}{3^n}/24]

    therefore 24Un = 24[7^(2n)] + [2^(3n)][3^n]

    now, Un+1 = 49[7^(2n)] + [2^(3n)][3^n]

    therefore Un+1 - 24Un = 25[7^(2n)]

    which is divisible by 25.

    Then follow the same argument as before

    ie U1 = 50 = 2x25
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    (Original post by Faaip De Oiad)
    similar method as before, Un = 7^(2n) + [2^(3n -3)][3^(n-1)]

    =7^(2n) + [{2^(3n)}{3^n}/24]

    therefore 24Un = 24[7^(2n)] + [2^(3n)][3^n]

    now, Un+1 = 49[7^(2n)] + [2^(3n)][3^n]

    therefore Un+1 - 24Un = 25[7^(2n)]

    which is divisible by 25.

    Then follow the same argument as before

    ie U1 = 50 = 2x25
    thank you thank you. i was wondering as to how to use them. and i can see why u are addicted. you are good at them :p:
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    98.
    (i)
    Let f(n) = 7^n - 6n - 1, and assume that f(k) is divisible by 36. Now:
    f(k+1) = 7^(k+1) - 6k - 7 = 7.7^k - 6k - 7
    And:
    f(k+1) - f(k) = (7-1)7^k - 6 = 6(7^k - 1), thus f(k+1) is divisible by 6 iff f(k) is divisible by 6. (Check it for n=2 to conclude the induction.)

    Now we want to prove that 7^k-1 is divisible by 6. Let g(n) = 7^n - 1, and assume as before that g(k) is divisible by 6. Now:
    g(k+1) = 7^(k+1) - 1
    And:
    g(k+1) - g(k) = 6(7^k), thus g(k+1) is divisible by 6 iff g(k) is divisible by 6. (Again, check it for n=2 to complete the induction.)

    Since f(k) = 6g(k), and g(k)=6m, then: f(k) = 6*6m = 36m. This concludes the proof.

    (ii) Similar argument as above, except show that both are divisible by 4 and hence the expression is divisible by 4*4=16.

    (iii)
    We know that:
    f(n) = 7^n - 6n - 1 is divisible by 36 (and hence by 4)
    h(n) = 5^n - 4n - 1 is divisible by 16 (and hence by 4)

    Now:
    f(n) - h(n) = 7^n - 5^n - 2n

    So: 7^n - 5^n - 2n is divisible by 4 since the LHS is divisible by 4. QED.
 
 
 
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