Calculate the value of the sum 1+3+5+....+2n+1. Use the mathematical induction to verify the result of the sum.

We observe that the terms of the sum are the terms of an
arithmetical progression. The ratio of the progression is r=2 and it's calculated using
the first and the second
term.

3-1=2

The number of
terms is n+1.

The sum of n+1 terms of an arithmetical
progression could be written:

Sn = (a1+a(n+1))*n/2, where
a1 is the first term of the progression and an is the last
term.

In our case, a1=1 and a(n+1)=(2n+1). By substituting
them into the formula of the sum, we'll obtain:

Sn =
(1+2n+1)*(n+1)/2

Sn =
(2n+2)(n+1)/2

After factorizing, we'll
get:

Sn = 2(n+1)(n+1)/2

Sn =
(n+1)^2

To demonstrate that the value of the sum is
(n+1)^2, we'll  use the method of mathematical induction, which consists in 3
steps:

1) verify that the method works for the number
1;

2) assume that the method works for an arbitrary number,
k;

3) prove that if the method works for an arbitrary
number k, then it work for the number k+1, too.

After the 3
steps were completed, then the formula works for any
number.

We'll start the first
step:

1) 1=1^2 => 1=1
true.

2) 1+3+5+...+(2k+1)=(k+1)^2 ,
true.

3) If 1+3+5+...+(2k+1)=(k+1)^2,
then

1+3+5+...+(2k+1)+(2k+2+1)=(k+2)^2

Let's
see if it is true.

We notice that the sum from the left
contains the assumed true sum, 1+3+5+...+(2k+1)=(k+1)^2. So, we'll re-write the sum by
substituting 1+3+5+...+(2k+1) with (k+1)^2.

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

We'll open the
brackets:

k^2 + 2k + 1 + 2k + 3 = k^2 + 4k + 4
true.

4) The 3 steps were completed, so the identity is
true for any value of n.

1+3+5+...+(2n+1) =
(n+1)^2