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📜  找出级数3,-6,12,-24的总和。 。 。最多N个词

📅  最后修改于: 2021-04-29 10:22:32             🧑  作者: Mango

给定整数N。任务是找到高达给出一系列的N项之和

例子

Input : N = 5
Output : Sum = 33

Input : N = 20
Output : Sum = -1048575

通过观察给定的级数,可以看出每一项与前一项的比率为-2 。因此,给定的级数是GP(几何级数)级数。
您可以在此处了解有关GP系列的更多信息。
所以, S_{n} = \frac{a(1-r^{n})}{1-r}   当r <0时。
在以上GP系列中,第一项i:e a = 3且公共比率i:e r =(-2)
所以, S_{n} = \frac{3(1-(-2)^{n})}{1-(-2)}
因此, S_{n} = 1-(-2)^{n}
下面是上述方法的实现:

C++
//C++ program to find sum upto N term of the series:
// 3, -6, 12, -24, .....
 
#include
#include
using namespace std;
//calculate sum upto N term of series
 
class gfg
{
    public:
    int Sum_upto_nth_Term(int n)
    {
        return (1 - pow(-2, n));
    }
};
// Driver code
int main()
{
    gfg g;
    int N = 5;
    cout<


Java
//Java program to find sum upto N term of the series:
// 3, -6, 12, -24, .....
 
import java.util.*;
//calculate sum upto N term of series
 
class solution
{
 
static int Sum_upto_nth_Term(int n)
{
    return (1 -(int)Math.pow(-2, n));
}
 
// Driver code
public static void main (String arr[])
{
    int N = 5;
    System.out.println(Sum_upto_nth_Term(N));
}
 
}


Python
# Python program to find sum upto N term of the series:
# 3, -6, 12, -24, .....
 
# calculate sum upto N term of series
def Sum_upto_nth_Term(n):
    return (1 - pow(-2, n))
 
# Driver code
N = 5
print(Sum_upto_nth_Term(N))


C#
// C# program to find sum upto
// N term of the series:
// 3, -6, 12, -24, .....
 
// calculate sum upto N term of series
class GFG
{
 
static int Sum_upto_nth_Term(int n)
{
    return (1 -(int)System.Math.Pow(-2, n));
}
 
// Driver code
public static void Main()
{
    int N = 5;
    System.Console.WriteLine(Sum_upto_nth_Term(N));
}
}
 
// This Code is contributed by mits


PHP


Javascript


输出:
33