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

*print*Print*list*Cite

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**

1+3+5+...+(2n+1) to calculate the sum.

Solution:

There are n+1 terms.

Let s = 1+3+5+..... (2n+1).......(1) Rearrnging the right from the last to first

s = 2n+1 +(2n-1)+1+....5+3+1.....(2)

Adding the two series term by term vertically:

2s = (2n+2) +(2n+2)+(2n+2)+....(2n+2). Or n+1 times 2n+2 . Or

2s = (2n+2)(n+1). Or

s = (2n+2)(n+1)/2 =( n+1)^2.

Induction proof:

1+3+5+.....2n+1 = (n+1)^2 be true for the n+1 terms

Adding the next n+2 nd term 2n+3 to both sides,

LHS: 1+3+5+..(2n+1)+(2n+3) .

RHS = (n+1)^2+(2n+3) = n^2+2n+1+2n+3 = n^2+4n+4 = (n+2)^2. which proves that the sum holds for n+2nd term. So it holds for all terms by induction.