A First Course in Linear Algebra, 2017a

A.2. Well Ordering and Induction 533
= 1
The sum on the left hand side also equals 1, so this equation is true for n = 1.
Now suppose this formula is valid for some n ≥ 1wheren is an integer. Hence, the following equation
is true.
n
∑ k 2 n(n + 1)(2n + 1)
= (1.1)
k=1
6
We want to show that this is true for n + 1.
Suppose we add (n + 1) 2 to both sides of equation 1.1.
n+1
∑ k 2 =
k=1
=
n
∑ k 2 +(n + 1) 2
k=1
n(n + 1)(2n + 1)
6
+(n + 1) 2
The step going from the first to the second line is based on the assumption that the formula is true for n.
Now simplify the expression in the second line,
This equals
and
Therefore,
n(2n + 1)
6
n(n + 1)(2n + 1)
6
+(n + 1) 2
( )
n(2n + 1)
(n + 1)
+(n + 1)
6
+(n + 1)= 6(n + 1)+2n2 + n
6
=
(n + 2)(2n + 3)
6
n+1
∑ k 2 (n + 1)(n + 2)(2n + 3) (n + 1)((n + 1)+1)(2(n + 1)+1)
= =
k=1
6
6
showing the formula holds for n + 1 whenever it holds for n. This proves the formula by mathematical
induction. In other words, this formula is true for all n = 1,2,···.
♠
Consider another example.
Example A.10: Proving an Inequality by Induction
Show that for all n ∈ N, 1 2 · 3 ···2n − 1 1
< √ .
4 2n 2n + 1
Solution. Again we will use the procedure given in Procedure A.8 to prove that this statement is true for
all n. Suppose n = 1. Then the statement says
which is true.
1
2 < √ 1
3