# First verify that Sn is valid for the first possible case by d9n1aQO

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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.

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.

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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