求数列 x(x+y)+x^2(x^2+y^2)+x^3(x^3+y^3)+…+x^n(x^n+y^n)的和
原文:https://www . geesforgeks . org/find-the-sum-the-series-xxy-x2 y2-x3x 3y 3-xnxnyn/
给定一系列
，其中 x、y 和 n 取整数值。任务是找到给定系列的第 n 个项的和。
示例:
Input: x = 2, y = 2, n = 2
Output: 40
Input: x = 2, y = 4, n = 2
Output: 92
进场:给定系列为:
。
这样我们的问题就简化为求两个 GP 系列的和。
以下是上述方法的实施:
C++
// CPP program to find the sum of series
#include <bits/stdc++.h>
using namespace std;
// Function to return required sum
int sum(int x, int y, int n)
{
// sum of first series
int sum1 = (pow(x, 2) * (pow(x, 2 * n) - 1))
/ (pow(x, 2) - 1);
// sum of second series
int sum2 = (x * y * (pow(x, n) * pow(y, n) - 1))
/ (x * y - 1);
return sum1 + sum2;
}
// Driver Code
int main()
{
int x = 2, y = 2, n = 2;
// function call to print sum
cout << sum(x, y, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the sum of series
public class GFG {
// Function to return required sum
static int sum(int x, int y, int n)
{
// sum of first series
int sum1 = (int) (( Math.pow(x, 2) * (Math.pow(x, 2 * n) - 1))
/ (Math.pow(x, 2) - 1));
// sum of second series
int sum2 = (int) ((x * y * (Math.pow(x, n) * Math.pow(y, n) - 1))
/ (x * y - 1));
return sum1 + sum2;
}
// Driver code
public static void main (String args[]){
int x = 2, y = 2, n = 2;
// function call to print sum
System.out.println(sum(x, y, n));
}
// This code is contributed by ANKITRAI1
}
Python 3
# Python3 program to find the sum of series
# Function to return required sum
def sum(x,y,n):
# sum of first series
sum1 = ((x**2)*(x**(2*n)-1))//(x**2 - 1)
# sum of second series
sum2 = (x*y*(x**n*y**n-1))//(x*y-1)
return (sum1+sum2)
# Driver Code
if __name__=='__main__':
x = 2
y = 2
n = 2
# function call to print sum
print(sum(x, y, n))
# this code is contributed by sahilshelangia
C
// C# program to find the sum of series
using System;
class GFG
{
// Function to return required sum
static int sum(int x, int y, int n)
{
// sum of first series
int sum1 = (int) ((Math.Pow(x, 2) *
(Math.Pow(x, 2 * n) - 1)) /
(Math.Pow(x, 2) - 1));
// sum of second series
int sum2 = (int) ((x * y * (Math.Pow(x, n) *
Math.Pow(y, n) - 1)) / (x * y - 1));
return sum1 + sum2;
}
// Driver code
public static void Main ()
{
int x = 2, y = 2, n = 2;
// function call to print sum
Console.Write(sum(x, y, n));
}
}
// This code is contributed by ChitraNayal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the
// sum of series
// Function to return required sum
function sum($x, $y, $n)
{
//sum of first series
$sum1 = (pow($x, 2) *
(pow($x, 2 * $n) - 1)) /
(pow($x, 2) - 1);
// sum of second series
$sum2 = ($x * $y * (pow($x, $n) *
pow($y, $n) - 1)) /
($x * $y - 1);
return $sum1 + $sum2;
}
// Driver code
$x = 2;
$y = 2;
$n = 2;
// function call to print sum
echo sum($x, $y, $n);
// This code is contributed
// by Shashank_Sharma
?>
java 描述语言
<script>
// java script program to find the
// sum of series
// Function to return required sum
function sum(x, y, n)
{
//sum of first series
sum1 = (Math.pow(x, 2) *
(Math.pow(x, 2 * n) - 1)) /
(Math.pow(x, 2) - 1);
// sum of second series
sum2 = (x * y * (Math.pow(x, n) *
Math.pow(y, n) - 1)) /
(x * y - 1);
return sum1 + sum2;
}
// Driver code
let x = 2;
let y = 2;
let n = 2;
// function call to print sum
document.write(sum(x, y, n));
// This code is contributed
// by bobby
</script>
Output:
40
时间复杂度: O(log(n))
版权属于：月萌API www.moonapi.com，转载请注明出处
