Dado el valor de n. Tienes que encontrar la suma de la serie donde el término n de la secuencia está dado por:
T n = n 2 – ( n – 1 ) 2
Ejemplos :
Input : 3
Output : 9
Explanation: So here the term of the sequence upto n = 3 are:
1, 3, 5 And hence the required sum is = 1 + 3 + 5 = 9
Input : 6
Output : 36
Enfoque simple
Simplemente use un ciclo y calcule la suma de cada término e imprima la suma.
C++
// CPP program to find summation of series
#include <bits/stdc++.h>
using namespace std;
int summingSeries(long n)
{
// use of loop to calculate
// sum of each term
int S = 0;
for (int i = 1; i <= n; i++)
S += i * i - (i - 1) * (i - 1);
return S;
}
// Driver Code
int main()
{
int n = 100;
cout << "The sum of n term is: "
<< summingSeries(n) << endl;
return 0;
}
Java
// JAVA program to find summation of series
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
class GFG
{
// function to calculate sum of series
static int summingSeries(long n)
{
// use of loop to calculate
// sum of each term
int S = 0;
for (i = 1; i <= n; i++)
S += i * i - (i - 1) * (i - 1);
return S;
}
// Driver code
public static void main(String[] args)
{
int n = 100;
System.out.println("The sum of n term is: " +
summingSeries(n));
}
}
Python3
# Python3 program to find summation
# of series
def summingSeries(n):
# use of loop to calculate
# sum of each term
S = 0
for i in range(1, n+1):
S += i * i - (i - 1) * (i - 1)
return S
# Driver Code
n = 100
print("The sum of n term is: ",
summingSeries(n), sep = "")
# This code is contributed by Smitha.
C#
// C# program to illustrate...
// Summation of series
using System;
class GFG
{
// function to calculate sum of series
static int summingSeries(long n)
{
// Using the pow function calculate
// the sum of the series
return (int)Math.Pow(n, 2);
}
// Driver code
public static void Main(String[] args)
{
int n = 100;
Console.Write("The sum of n term is: " +
summingSeries(n));
}
}
// This code contribute by Parashar...
PHP
<?php
// PHP program to find
// summation of series
function summingSeries( $n)
{
// use of loop to calculate
// sum of each term
$S = 0;
for ($i = 1; $i <= $n; $i++)
$S += $i * $i - ($i - 1) *
($i - 1);
return $S;
}
// Driver Code
$n = 100;
echo "The sum of n term is: ",
summingSeries($n) ;
// This code contribute by vt_m.
?>
Javascript
<script>
// Javascript program to find summation of series
function summingSeries(n)
{
// use of loop to calculate
// sum of each term
let S = 0;
for (let i = 1; i <= n; i++)
S += i * i - (i - 1) * (i - 1);
return S;
}
// Driver Code
let n = 100;
document.write("The sum of n term is: " + summingSeries(n));
// This code is contributed by rishavmahato348.
</script>
Producción:
The sum of n term is: 10000
Complejidad de tiempo – O(N)
Complejidad de espacio – O(1)
Enfoque eficiente
El uso del enfoque matemático puede resolver este problema de manera más eficiente.
T n = n 2 – (n-1) 2
La suma de la serie viene dada por (S) = SUM( T n )
TOMEMOS UN EJEMPLO SI
N = 4
Significa que debe haber 4 términos en la serie por lo que
1 er término = 1 2 – ( 1 – 1 ) 2
2 do término = 2 2 – ( 2 – 1 ) 2
3 o término = 3 2 – ( 3 – 1 ) 2
4 o término = 4 2 – ( 3 – 1 ) 2
ASÍ QUE LA SUMA ESTÁ DADA POR = (1 – 0) + (4 – 1) + (9 – 4) + (16 – 9)
= 16
DE ESTO TENEMOS AVISO DE QUE 1, 4, 9 SE CANCELAN DE LA SERIE
SOLO QUEDA 16 QUE ES IGUAL AL CUADRADO DE N
Entonces, de la serie anterior notamos que cada término se cancela del siguiente término, solo el último término es izquierda que es igual a N 2 .
C++
// CPP program to illustrate...
// Summation of series
#include <bits/stdc++.h>
using namespace std;
int summingSeries(long n)
{
// Sum of n terms is n^2
return pow(n, 2);
}
// Driver Code
int main()
{
int n = 100;
cout << "The sum of n term is: "
<< summingSeries(n) << endl;
return 0;
}
Java
// JAVA program to illustrate...
// Summation of series
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
class GFG
{
// function to calculate sum of series
static int summingSeries(long n)
{
// Using the pow function calculate
// the sum of the series
return (int)Math.pow(n, 2);
}
// Driver code
public static void main(String[] args)
{
int n = 100;
System.out.println("The sum of n term is: " +
summingSeries(n));
}
}
Python3
# Python3 program to illustrate...
# Summation of series
import math
def summingSeries(n):
# Sum of n terms is n^2
return math.pow(n, 2)
# Driver Code
n = 100
print ("The sum of n term is: ",
summingSeries(n))
# This code is contributed by mits.
C#
// C# program to illustrate...
// Summation of series
using System;
class GFG
{
// function to calculate sum of series
static int summingSeries(long n)
{
// Using the pow function calculate
// the sum of the series
return (int)Math.Pow(n, 2);
}
// Driver code
public static void Main()
{
int n = 100;
Console.Write("The sum of n term is: " +
summingSeries(n));
}
}
// This code is contributed by nitin mittal.
PHP
<?php
// PHP program to illustrate...
// Summation of series
function summingSeries($n)
{
// Sum of n terms is n^2
return pow($n, 2);
}
// Driver Code
$n = 100;
echo "The sum of n term is: ",
summingSeries($n);
// This code contribute by vt_m.
?>
Javascript
<script>
// Javascript program to illustrate...
// Summation of series
function summingSeries(n)
{
// Sum of n terms is n^2
return Math.pow(n, 2);
}
// Driver Code
let n = 100;
document.write("The sum of n term is: "
+ summingSeries(n) + "<br>");
// This code is contributed by subham348.
</script>
Producción:
The sum of n term is: 10000
Complejidad temporal – O(1)
Complejidad espacial – O(1)
Publicación traducida automáticamente
Artículo escrito por Kanishk_Verma y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA