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.
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
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.
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.
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.