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```Mathematical Induction (10 points)
To prove a theorem holds true for every natural number, n, you can use induction.
1. Prove it is true for n = 1 (or some specific number.) This is called the “base” case.
2. Assuming the theorem is true for n = k (this is called the “induction hypothesis”.)
3. Now do the work to show the next case follows. That is, if the theorem is true for n = k,
then it is also true for n = k+1.
Here is a website that I found that might help:
http://www.themathpage.com/aprecalc/mathematical-induction.htm
 n(n + 1) 
They have a nice example proving that 13 + 2 3 + ... + n 3 = 
and another proving that
 2 
the sum of the first n odd numbers is equal to the square of n.
2
Here are 2 problems for you to do:
1. (4 points) Prove that:
1 ⋅ 3 + 2 ⋅ 4 + 3 ⋅ 5 + ... + n(n + 2) =
n(n + 1)(2n + 7)
6
2. (6 points) Prove by using mathematical induction: For any positive integer n,
5
6
n+4
n(3n + 7)
+
+...+
=
1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4
n(n + 1)(n + 2) 2(n + 1)(n + 2)
Here are some nice formulas:
Sigma Notation = Closed Form
Expanded
1 + 1 + 1 + ... + 1 (n times)
1 + 2 + 3 + ... + n
1 + 4 + 9 + ... + n2
1 + 8 + 27 + ... + n3
1 + 16 + 81 + ... + n4
1 + 32 + 243 + ... + n5
```
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