MI 4
Name:
Mathematical Induction
Let’s look at another example, this time something we’ve already seen before. There are
often many different ways to prove the same result; here, we will prove that
n
X
k=
k=1
n(n + 1)
,
2
n≥1
by mathematical induction.
The base case is relatively easy. Note that for n = 1,
1
X
k=1=
k=1
1(2)
.
2
Proceeding to the induction step, we first assume that
n
X
k=
k=1
We want to show that
n+1
X
k=
k=1
n(n + 1)
.
2
(n + 1)(n + 2)
.
2
But we have
n+1
X
k=
n
X
k + (n + 1)
k=1
k=1
n(n + 1)
+n+1
2 n
= (n + 1)
+1
2
(n + 1)(n + 2)
=
.
2
=
by assumption
Then by mathematical induction, we see that
n
X
k=1
k=
n(n + 1)
2
for all n ≥ 1.
Induction 1.1
Spring 2010
MI 4
Name:
Now here are some to try on your own.
1. Prove that
n
X
k=1
k2 =
n(n + 1)(2n + 1)
.
6
2. Find the sum of the first n odd integers, beginning at 1, and prove your result using
induction.
3. Consider the arithmetic sequence ak = 3k − 2. Using what you have already learned,
find the sum
n
X
ak .
k=1
Then prove your result is valid using induction.
4. Using what you have already learned, find
n
X
2k .
k=0
Then prove this result is true using induction.
Induction 1.2
Spring 2010