1) Mathematical induction is a method of proof that can be used to prove statements for all positive integers. It involves showing that a statement is true for n=1, and assuming it is true for an integer k to prove it is true for k+1. 2) The document provides an example using mathematical induction to prove the formula Sn = n(n+1) for the sum of the first n even integers. 3) Finite differences are used to determine if a sequence has a quadratic model by seeing if the second differences are constant. The example finds the quadratic model n^2 for the sequence 1, 4, 9, 16, 25, 36.

1. Mathematical
Induction
Digital Lesson

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Mathematical induction is a legitimate method of
proof for all positive integers n.
Principle:
Let Pn be a statement involving n, a positive integer.
If
1. P1 is true, and
2. the truth of Pk implies the truth of Pk + 1 for
every positive k,
then Pn must be true for all positive integers n.

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Example:
Find Pk + 1 for
3(2 1)
: .
1k k
k
P S
k
+
=
−
1 1
3[2( 1)
11
1]
:k kP S
k
k+ +
+ +
+
=
−
Replace k by k + 1.
Simplify.
3(2 2 1)k
k
+ +
=
3(2 3)k
k
+
= Simplify.

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Example:
Use mathematical induction to prove
Sn = 2 + 4 + 6 + 8 + . . .
+ 2n = n(n + 1)
for every positive integer n.
1. Show that the formula is true when n = 1.
S1 = n(n + 1) = 1(1 + 1) = 2 True
2. Assume the formula is valid for some integer k.
Use this assumption to prove the formula is valid
for the next integer, k + 1 and show that the
formula Sk + 1 = (k + 1)(k + 2) is true.
Sk = 2 + 4 + 6 + 8 + . . .
+ 2k = k(k + 1) Assumption
Example continues.

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Example continued:
Sk + 1 = 2 + 4 + 6 + 8 + . . .
+ 2k + [2(k + 1)]
= 2 + 4 + 6 + 8 + . . .
+ 2k + (2k + 2)
= Sk + (2k + 2) Group terms to form Sk.
= k(k + 1) + (2k + 2) Replace Sk by k(k + 1).
= k2
+ k + 2k + 2 Simplify.
= k2
+ 3k + 2
= (k + 1)(k + 2)
The formula Sn = n(n + 1) is valid for all positive integer
values of n.
= (k + 1)((k + 1)+1)

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Sums of Powers of Integers :
1
( 1)
1 2 3 41
2
.
n
i
n n
i n
=
+
= + + + + + =∑ L
2 2 2 2 2 2
1
( 1)(2 1)
1 2 3 42.
6
n
i
n n n
i n
=
+ +
= + + + + + =∑ L
2 2
3 3 3 3 3 3
1
( 1)
1 2 3 43.
4
n
i
n n
i n
=
+
= + + + + + =∑ L
2
4 4 4 4 4 4
1
( 1)(2 1)(3 3 1)
1 2 3 44.
30
n
i
n n n n n
i n
=
+ + + −
= + + + + + =∑ L
2 2 2
5 5 5 5 5 5
1
( 1) (2 2 1)
1 2 3 4
1
5
2
.
n
i
n n n n
i n
=
+ + −
= + + + + + =∑ L

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2 2 2 2 2 2
1
( 1)(2 1)
1 2 3 4 .
6
n
i
n n n
i n
=
+ +
= + + + + + =∑ L
Example:
Use mathematical induction to prove for all positive integers n,
Assumption2 2 2 2 2 ( 1)(2 1)
1 2 3 4
6k kS
k k k+ +
= + + + + + =L
2 2 2
1
2 2 2
1 2 3 4 ( 1)k k kS + = + + + + ++ + L
2
( 1)kS k= + +
2
2 1kS k k= + + +
2( 1)(2 1)
2 1
6
k k k
k k
+ +
= + + +
Group terms to form Sk.
Replace Sk by k(k + 1).
Example continues.
1
( 1)(2( ) 1) 1(2)(2 1) 6 1
6
11
6 6
1
S
+ + +
= = = = True

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3 2 2
2 3 6 12 6
6 6
k k k k k+ + + += + Simplify.
Example continued:
3 2
2 9 13 6
6
k k k+ + +=
2
( 3 2)(2 3)
6
k k k+ + +
=
( 1)( 2)(2 3)
6
k k k+ + +
=
( )[( ) 1][2(1 ) ]
6
1 11k k k+ + ++ +
=
The formula is valid for all positive
integer values of n.
( 1)(2 1)
6n
n n n
S
+ +
=

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Finite Differences
The first differences of the sequence 1, 4, 9, 16, 25, 36 are
found by subtracting consecutive terms.
n: 1 2 3 4 5 6
an: 1 4 9 16 25 36
First differences: 3 5 7 9 11
Second differences: 2 2 2 2
The second differences are found by subtracting consecutive
first differences.
quadratic model

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When the second differences are all the same nonzero number,
the sequence has a perfect quadratic model.
Find the quadratic model for the sequence
1, 4, 9, 16, 25, 36, . . .
an = an2
+ bn + c
a1 = a(1)2
+ b(1) + c = 1
a2 = a(2)2
+ b(2) + c = 4
a3 = a(3)2
+ b(3) + c = 9
Solving the system yields a = 1, b = 0, and c = 0.
an = n2
a + b + c = 1
4a + 2b + c = 4
9a + 3b + c = 9

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an = an2
+ bn + c
a0 = a(0)2
+ b(0) + c = 3
a1 = a(1)2
+ b(1) + c = 3
a4 = a(4)2
+ b(4) + c = 15
an = n2
– n + 3
c = 3
a + b + c = 3
16a + 4b + c = 15
Solving the system yields
a = 1, b = –1, and c = 3.
Example:
Find the quadratic model for the sequence with
a0 = 3, a1 = 3, a4 = 15.