Dada la secuencia de números impares
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23,….
Encuentra la suma de los primeros n números impares
Ejemplos:
Input : n = 2 Output : 4 Sum of first two odd numbers is 1 + 3 = 4. Input : 5 Output : 25 Sum of first 5 odd numbers is 1 + 3 + 5 + 7 + 9 = 25
Una solución simple es iterar a través de todos los números impares.
C++
// A naive CPP program to find sum of
// first n odd numbers
#include <iostream>
using namespace std;
// Returns the sum of first
// n odd numbers
int oddSum(int n)
{
int sum = 0, curr = 1;
for (int i = 0; i < n; i++) {
sum += curr;
curr += 2;
}
return sum;
}
// Driver function
int main()
{
int n = 20;
cout << " Sum of first " << n
<< " Odd Numbers is: " << oddSum(n);
return 0;
}
Java
// Java program to find sum of
// first n odd numbers
import java.util.*;
class Odd
{
// Returns the sum of first
// n odd numbers
public static int oddSum(int n)
{
int sum = 0, curr = 1;
for (int i = 0; i < n; i++) {
sum += curr;
curr += 2;
}
return sum;
}
// driver function
public static void main(String[] args)
{
int n = 20;
System.out.println(" Sum of first "+ n
+" Odd Numbers is: "+oddSum(n));
}
}
// This code is contributed by rishabh_jain
Python3
# Python3 program to find sum
# of first n odd numbers
def oddSum(n) :
sum = 0
curr = 1
i = 0
while i < n:
sum = sum + curr
curr = curr + 2
i = i + 1
return sum
# Driver Code
n = 20
print (" Sum of first" , n, "Odd Numbers is: ",
oddSum(n) )
# This code is contributed by rishabh_jain
C#
// C# program to find sum of
// first n odd numbers
using System;
class GFG {
// Returns the sum of first
// n odd numbers
public static int oddSum(int n)
{
int sum = 0, curr = 1;
for (int i = 0; i < n; i++) {
sum += curr;
curr += 2;
}
return sum;
}
// driver function
public static void Main()
{
int n = 20;
Console.WriteLine(" Sum of first " + n
+ " Odd Numbers is: " + oddSum(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// A naive PHP program to find sum of
// first n odd numbers
// Returns the sum of first
// n odd numbers
function oddSum($n)
{
$sum = 0; $curr = 1;
for ($i = 0; $i < $n; $i++)
{
$sum += $curr;
$curr += 2;
}
return $sum;
}
// Driver Code
$n = 20;
echo " Sum of first ", $n
, " Odd Numbers is: ", oddSum($n);
// This code is contributed by vt_m.
?>
Javascript
<script>
// A naive Javascript program to find sum of
// first n odd numbers
// Returns the sum of first
// n odd numbers
function oddSum(n)
{
let sum = 0; curr = 1;
for (let i = 0; i < n; i++)
{
sum += curr;
curr += 2;
}
return sum;
}
// Driver Code
let n = 20;
document.write(" Sum of first " + n
+ " Odd Numbers is: " + oddSum(n));
// This code is contributed by gfgking.
</script>
Producción:
Sum of first 20 odd numbers is 400
Complejidad de Tiempo: O(n)
Espacio Auxiliar: O(1)
Una solución eficiente es usar la fórmula directa. Para encontrar la suma de los primeros n números impares, podemos aplicar el teorema de los números impares, establece que la suma de los primeros n números impares es igual al cuadrado de n.
∑(2i – 1) = n 2 donde i varía de 1 a n
Sea n = 10, por lo tanto, la suma de los 10 primeros números impares es
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
si aplicamos el teorema de los números impares:
suma de los 10 primeros números impares = n * n = 10 * 10 = 100.
A continuación se muestra la implementación del enfoque anterior:
C++
// Efficient program to find sum of
// first n odd numbers
#include <iostream>
using namespace std;
// Returns the sum of first
// n odd numbers
int oddSum(int n)
{
return (n * n);
}
// Driver function
int main()
{
int n = 20;
cout << " Sum of first " << n
<< " Odd Numbers is: " << oddSum(n);
return 0;
}
Java
// Java program to find sum of
// first n odd numbers
import java.util.*;
class Odd
{
// Returns the sum of first
// n odd numbers
public static int oddSum(int n)
{
return (n * n);
}
// driver function
public static void main(String[] args)
{
int n = 20;
System.out.println(" Sum of first "+ n
+" Odd Numbers is: "+oddSum(n));
}
}
// This code is contributed by rishabh_jain
Python3
# Python3 program to find sum
# of first n odd numbers
def oddSum(n) :
return (n * n);
# Driver Code
n = 20
print (" Sum of first" , n, "Odd Numbers is: ",
oddSum(n) )
# This code is contributed by rishabh_jain
C#
// C# program to find sum of
// first n odd numbers
using System;
class GFG {
// Returns the sum of first
// n odd numbers
public static int oddSum(int n)
{
return (n * n);
}
// driver function
public static void Main()
{
int n = 20;
Console.WriteLine(" Sum of first " + n
+ " Odd Numbers is: " + oddSum(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// Efficient program to find sum of
// first n odd numbers
// Returns the sum of first
// n odd numbers
function oddSum($n)
{
return ($n * $n);
}
// Driver Code
$n = 20;
echo " Sum of first " , $n,
" Odd Numbers is: ", oddSum($n);
// This code is contributed by vt_m.
?>
Javascript
<script>
// Javascript program to find sum of first n odd numbers
// Returns the sum of first
// n odd numbers
function oddSum(n)
{
return (n * n);
}
let n = 20;
document.write(" Sum of first " + n
+ " Odd Numbers is: " + oddSum(n));
// This code is contributed by divyesh072019.
</script>
Producción:
Sum of first 20 odd numbers is 400
Complejidad Temporal: O(1)
Espacio Auxiliar : O(1)
¿Cómo funciona?
Podemos demostrarlo usando inducción matemática. Sabemos que es cierto para n = 1 y n = 2 ya que las sumas son 1 y 4 (1 + 3) respectivamente.
Let it be true for n = k-1. Sum of first k odd numbers = Sum of first k-1 odd numbers + k'th odd number = (k-1)*(k-1) + (2k - 1) = k*k
Publicación traducida automáticamente
Artículo escrito por ARSHPREET_SINGH y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA