- a form of mathematical proof

Lets look at the patterns:

S

_{4}= 1 + 2 + 3 + 4 = 10

S

_{6}= 1 + 2 + 3 + 4 + 5 + 6 = 21

S

_{8}= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

S

_{10}= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55

When n = 4, S

_{4}= 10, what are the factors of 10? (2)(5)

When n = 6, S

_{6}= 21, what are the factors of 21? (3)(7)

When n = 8, S

_{8}= 36, what are the factors of 36 that fit the pattern? (4)(9)

When n = 10, S

_{10}= 55, and again, the factors are (5)(11)

therefore we can conclude that S

_{n}= (n/2)(n + 1)

Now lets try another one:

S

_{1}= 1 = 1

^{2}

S

_{2}= 1 + 3 = 2

^{2}

S

_{3}= 1 + 3 + 5 = 3

^{2}

S

_{4}= 1 + 3 + 5 + 7 = 4

^{2}

S

_{5}= 1 + 3 + 5 + 7 + 9 = 5

^{2}

S

_{n}= 1 + 3 + 5 + 7 + ... + (2n - 1) = n

^{2}

Looking at the pattern we see that 2n - 1 represents consecutive odd integers and n

^{2}represents the sum of the odd integers.

By looking at patterns, sometimes we may come up with a wrong conclusion:

Example:

Take a circle and put a dot on it. How many areas do you have? 1

Now put 2 dots on it and connect the dots. How many areas do you have? 2

Now put 3 total dots on it and connect the dots. How many areas do you have? 4

Now put 4 total dots on it and connect the dots. How many areas do you have? 8

Now put 5 total dots on it and connect the dots. How many areas do you have? 16

Lets look at what we have seen -

1 to 1

2 to 2

3 to 4

4 to 8

5 to 16

so what do you conclude 6 would go to?

Normally we would say 32, but it actually has 30 or 31 depending if the dots are equidistant from each other. So this pattern does not work.

Therefore we have to prove a formula works for all numbers, you have to do the following.

The Principle of Mathematical Induction:

Let P

_{n}be a statement involving the positive integer n. If:

1. P

_{1}is true AND

2. The truth of P

_{k}implies the truth of P

_{k+1}, for every positive k, then P

_{n}must be true for all positive integers n.

Example:

So S

_{n}= 1 + 3 + 5 + 7 + ... + (2n - 1) = n

^{2}

Let n = 1

1. S

_{1}= 2(1) - 1 = 1 and 1

^{2}= 1 both are true

2. Then Assume S

_{k}= 1 + 3 + 5 + 7 + ... + (2k - 1) = k

^{2}

Then S

_{k+1}= 1 + 3 + 5 + 7 + ... + (2k - 1) + (2(k + 1) - 1 ) = (k + 1)

^{2}

S

_{k+1}= S

_{k}+ (2k + 2 - 1 ) = (k + 1)

^{2}

k

^{2}+ 2k + 1 = (k + 1)

^{2}

(k + 1)

^{2}= (k + 1)

^{2}

this is what we had to prove so

Therefore this formula is valid for all positive integer values for n.

Example 2: Find P

_{k+1}for the given P

_{k}

P

_{k}= (k/2)(5k - 3)

P

_{k+1}= ((k + 1)/2)(5(k+1) - 3)

= ((k + 1)/2)(5k + 5 - 3)

= ((k + 1)/2)(5k + 2)

= (1/2)(k + 1)(5k + 2)

= (1/2)(5k

^{2}+ 7k + 2)

Example 3:

S

_{n}= 1 + 5 + 9 + 13 + ... + (4n - 3) = n (2n - 1)

Let n = 1

S

_{1}= 4(1) - 3 = 1 and (1)(2(1) - 1) = (1)(1) = 1 therefore it is true for both

Assume:

S

_{k}= 1 + 5 + 9 + 13 + ... + (4k - 3) = k (2k - 1)

then S

_{k+1}= 1 + 5 + 9 + 13 + ... + (4k - 3) + (4(k + 1) - 3) = (k + 1)(2 (k + 1) - 1)

S

_{k+1}= S

_{k}+ (4k + 4 - 3) = (k + 1)(2 k + 2 - 1)

S

_{k+1}= k (2k - 1) + (4k + 1) = (k + 1)(2 k + 1)

S

_{k+1}= 2k

^{2}- k + 4k + 1 = (k + 1)(2 k + 1)

S

_{k+1}= 2k

^{2}+ 3k + 1 = (k + 1)(2 k + 1)

S

_{k+1}=(k + 1)(2 k + 1) = (k + 1)(2 k + 1)

Therefore the formula is valid for all positive integer values of n.

Example 4: 2 (1 + 3 + 3

^{2}+ 3

^{3}+ ... + 3

^{n -1}) = 3

^{n}- 1

S

_{n}= 2 (1 + 3 + 3

^{2}+ 3

^{3}+ ... + 3

^{n -1}) = 3

^{n}- 1

Let n = 1

S

_{1}= 2 (3

^{1-1}) = 2(3

^{0}) = 2(1) = 2 and 3

^{1}- 1 = 3 - 1 = 2

Assume S

_{k}= 2 (1 + 3 + 3

^{2}+ 3

^{3}+ ... + 3

^{k -1}) = 3

^{k}- 1

S

_{k+1}= 2 (1 + 3 + 3

^{2}+ 3

^{3}+ ... + 3

^{k -1}+ 3

^{(k+1)-1})= 3

^{k+1}- 1

S

_{k+1}= 2 (1 + 3 + 3

^{2}+ 3

^{3}+ ... + 3

^{k -1}+ 3

^{k}) = 3

^{k+1}- 1

S

_{k+1}= 2 (1 + 3 + 3

^{2}+ 3

^{3}+ ... + 3

^{k -1}) + 2(3

^{k}) = 3

^{k+1}- 1

S

_{k+1}= S

_{k}+ 2(3

^{k}) = 3

^{k+1}- 1

S

_{k+1}= 3

^{k}- 1 + 2(3

^{k}) = 3

^{k+1}- 1

S

_{k+1}= 3(3

^{k}) - 1 = 3

^{k+1}- 1

S

_{k+1}= 3

^{k+1}- 1 = 3

^{k+1}- 1

We can conclude that this formula holds for all positive integer values of n.

Sums of Powers of Integers - find on

http://www.libraryofmath.com/summation-formulas.html

or in textbook page 652

Finding a quadratic model

a

_{0}= 3, a

_{1}= 3, a

_{4}= 15, a

_{n}= an

^{2}+ bn + c

a

_{0}= a(0)

^{2}+ b(0) + c = 3 therefore c = 3

a

_{1}= a(1)

^{2}+ b(1) + c = 3

= a + b + 3 = 3

= a + b = 0

a = -b

a

_{4}= a(4)

^{2}+ b(4) + c = 15

= 16a + 4b + 3 = 15

= 16a + 4b = 12

16(-b) + 4b = 12

-12b = 12

b = -1

a = - b

a = -(-1) = 1

therefore a

_{n}= n

^{2}- n + 3

To check your answer, use the TI-83 calculator, put the data in:

L1 (0, 1, 4) and L2 (3, 3, 15)

Stat go to Quad Reg L1, L2

this gives you an = n

^{2}- n + 3

same answer!

_{}