Question

Use induction to prove the following:

- Prove that the sum of the first n odd integers is n2

writing a proof and then a program to go along with it.

Answer 1

Program

#include <stdio.h>
#include <math.h>
int main()
{
    int i, num, sum=0, sqr=0;
  
    printf("Enter Value Of N: ");
    scanf("%d", &num);
    i=num;
    while(i>0)
    {
       sum = sum + ((i * 2)-1);
       i--;  
    }
    printf("\nSum of First %d odd numbers is = %d\n",num ,sum);
   sqr = num*num;
   printf("\nSquare Root of %d is = %d\n",num, sqr);
    return 0;
}

OUTPUT

We need to Prove the following

1+3+5+...+(2n-1) =n²

1=1=1²
1+3=4=2²
1+3+5=9=3²
1+3+5+7=16=4²
1+3+5+7+9=25=5²

So the Answer is 1+3+5+...+(2n-1) = n²

Proof

Step 1.
For n = 1 it's true that 1 = 2*1-1

Step 2.
We suppose that 1+3+5+...+(2n-1) = n² and need to prove the equation 1+3+5+...+(2(n+1)-1) = (n+1)²

Hence if we add (2(n+1) -1) to the equation it becomes
1+3+5+...+(2n-1) = n²

And the equation becomes
1+3+5+...+(2n-1) + (2(n+1) -1) = n² + (2(n+1) -1)

After Simplification
1+3+5+...+(2(n+1) -1) = n² + 2n+2 -1

but we have n²+2n+1 = (n+1)²

So it proves our theorm.
1+3+5+...+(2(n+1)-1) = (n+1)²