First verify that Sn is valid for the first possible case

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					Advanced Mathematical Concepts                                                               Chapter 12

Lesson 12-9

Example 1
                                          n(3n - 1)
Prove Sn: 1 + 4 + 7 + … + (3n - 2) =               .
                                             2

1. First, verify that Sn is valid for n = 1.

                       1[3(1) - 1]
    Since S1 = 1 and       2       = 1, the formula is valid for n = 1.

2. Then, assume that Sn is valid for n = k.

                                        k(3k - 1)
    Sk  1 + 4 + 7 + ... + (3k - 2) =      2           Replace n with k.

    Next, prove that it is also valid for n = k + 1.

                                                         k(3k - 1)
    Sk + 1  1 + 4 + 7 + ... + (3k - 2) + (3k + 1) =          2     + (3k + 1)     3(k + 1) - 2 = 3k + 1
                                                         k(3k - 1) 2(3k + 1)
                                                       =            +
                                                              2           2
                                                            2
                                                         3k - k + 6k + 2
                                                       =
                                                                  2
                                                         3k2 + 5k + 2
                                                       =        2
                                                         (k + 1)(3k + 2)
                                                       =          2

    If k + 1 is substituted into the original formula, the same result is obtained. Thus, if the formula is
    valid for n = k, it is also valid for n = k + 1. Since Sn is valid for n = 1, it is also valid for n = 2, n = 3,
    and so on.
Advanced Mathematical Concepts                                                              Chapter 12

Example 2
Prove that 8n – 1 is divisible by 7 for all positive integers n.

Using the definition of divisibility, we can state the conjecture as follows.

Sn  8n - 1 = 7r for some integer r.

1. First verify that this is valid for n = 1.

    S1  81 - 1 = 7         Since 7 is divisible by 7, this is valid for n = 1.

2. Then assume that Sn is valid for n = k and use this assumption to prove that it is also valid for n = k +
   1.

    Sk  8k - 1 = 7r for some integer r.                Assume sk is true.
    Sk + 1  8k + 1 - 1 = 7t for some integer t.        Show that Sk + 1 must follow.

    For this proof, rewrite the left-hand side of Sk so that it matches the left-hand side of Sk + 1.

        8k - 1    = 7r           Sk
     8(8k - 1)    = 8(7r)        Multiply each side by 8.
     8k + 1 - 8   = 56r          Simplify.
     8k + 1 - 1   = 56r + 7      Add 7 to each side.
     8k + 1 - 1   = 7(8r + 1)    Factor.

    Let t = 8r + 1, an integer. Then 8k + 1 - 1 = 7t.

    We have shown that if Sk is valid, then Sk + 1 is also valid. Since Sn is valid for n = 1, it is also valid for
    n = 2, n = 3, and so on. Hence, 8n – 1 is divisible by 7 for all positive integers n.
Advanced Mathematical Concepts                                                               Chapter 12

Example 3
Prove Sn: 1 + 3 + 5 + 7 + … + (2n – 1) = n2.

1. First verify that Sn is valid for the first possible case, n = 1.

    If n = 1, then S1 is 1. The sum is 1 and 12 = 1, so the formula is valid for the first case.

2. Then assume that the formula is valid for n = k and prove that it is also valid for n = k + 1.

    It must be proven that if Sk: 1 + 3 + 5 + 7 + … = k2, then Sk + 1: 1 + 3 + 5 + 7 + … = (k + 1)2 is also
    valid.

    The kth term of the series is 2k – 1. Substitute k + 1 for k. Then 2(k + 1) – 1 = 2k + 1. You can write
    the formula for Sk+1 by adding the next term, 2k + 1, to each side.

    Sk  1 + 3 + 5 + 7 + ... + (2k - 1) = k2
    Sk + 1  1 + 3 + 5 + 7 + ... + (2k - 1) + (2k + 1) = k2 + (2k + 1)
                                                       = (k + 1)2

    Since the formula for the sum for n = k + 1 produces the same result as the direct computation of the
    sum of the series, the formula is valid for n = k + 1 if it is valid for n = k. Thus, it can be concluded
    that since the formula is valid for n = 1, it is also valid for n = 2. Since it is valid for n = 2, it is valid
    for n = 3, and so on, indefinitely. Therefore, the formula is valid for any positive integer n.

				
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