Write a C, C++ Program to find sum of first n odd numbers.

There are multiple approach to solve this problem. Here i use mathematical trick to solve this problem in efficient way.

An odd number is a number which is not a multiple of two. For example - 1 , 3 , 5 etc.

Program to check Even or Odd number.

Print Even Numbers between 1 to 100.

Let's say we have to calculate the sum of first 10 odd numbers.

1 , 3, 5, 7, 9, 11, 13, 15, 17, 19

Check the sequence of first 10 odd numbers it's an arithmetic progression with common difference 2. The formula of arithmetic progression is

Enter the number : 20

Sum of first 20 odd numbers are 400

The sum of first 100 odd numbers is : 10000

We already proved that

Find cube root of a number.

Reverse a string using stack.

There are multiple approach to solve this problem. Here i use mathematical trick to solve this problem in efficient way.

An odd number is a number which is not a multiple of two. For example - 1 , 3 , 5 etc.

Program to check Even or Odd number.

Print Even Numbers between 1 to 100.

## How to Calculate Sum of First n Odd Numbers

Let's say we have to calculate the sum of first 10 odd numbers.

**First 10 odd numbers are**1 , 3, 5, 7, 9, 11, 13, 15, 17, 19

Check the sequence of first 10 odd numbers it's an arithmetic progression with common difference 2. The formula of arithmetic progression is

So we can conclude that the

**sum of first n odd numbers is square of n**.## C Program to Find Sum of first n odd numbers

#include <stdio.h> int main(void) { int n,sum; printf("Enter the number"); scanf("%d",&n); sum = n*n; printf("Sum of first %d odd numbers are %d",n,sum); return 0; }

**Output :**Enter the number : 20

Sum of first 20 odd numbers are 400

## Find Sum of First 100 Odd Numbers

**METHOD 1:**

#include <iostream> using namespace std; int main() { int i=0, num=0, sum=0; while (i < 100) { /* Check if it's not divisible of 2 */ if(num%2 != 0) { /* Add number and increment the value of i */ sum = sum + num; i++; } num++; } cout << "The sum of first 100 odd numbers is " << sum; return 0; }

**Output :**The sum of first 100 odd numbers is : 10000

**METHOD 2:**

We already proved that

**sum of first n odd numbers is square of n**. Using this approach we can directly print the square of n where n is 100.Find cube root of a number.

Reverse a string using stack.

C++, C Plus Plus, Programlama

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