求系列 1 的 N 项之和， (1+4) ， (1+4+4^2)， (1+4+4^2+4^3), .....

给定一个正整数N 。找到系列的前 N 项的总和-

1, (1+4), (1+4+4²), (1+4+4²+4³), …., till N terms

例子：

Input: N = 3
Output: 27

Input: N = 5
Output: 453

方法：

1st term = 1

2nd term = (1 + 4)

3rd term = (1 + 4 + 4 ^ 2)

4th term = (1 + 4 + 4 ^ 2 + 4 ^ 3)

.

.

Nth term = (1 + 4 + 4 ^ 2+….+ 4 ^ (N – 2) + 4 ^(N – 1))

该序列是通过使用以下模式形成的。对于任何值 N-

$S_{N}=\frac{4}{9}(4^{N}-1)-\frac{N}{3}$

推导：

以下一系列步骤可用于推导公式以找到 N 项的总和 -

The series 

$1, (1+4), (1+4+4^{2}), (1+4+4^{2}+4^{3})+....+N terms$

can be decomposed as-

$a_{1}=1$

$a_{2}=1+4$

$a_{3}=1+4+4^{2}$

$a_{4}=1+4+4^{2}+4^{3}$

$a_{N}=1+4+4^{2}+4^{3}+....+4^{N}$                               -(1)

The equation (1) is in G.P. with

First term a = 1

Common ration r = 4

The sum of N terms in G.P. for r>1 is

$S_{N}=\frac{a(r^{N}-1)}{r-1}$

Substituting the values of a and r in the above equation, we get-

$S_{N}=\frac{1(4^{N}-1)}{4-1}$

Thus, the term

$a_{N}=\frac{(4^{N}-1)}{3}$

The sum of the series 1, (1+4), (1+4+4^{2}), (1+4+4^{2}+4^{3})+….+N terms can be represented as-

$S_{N}=\sum a^{N}$

$S_{N}=\sum \frac{4^{N}-1}{3}$

$S_{N}=\frac{1}{3}\sum 4^{N}-\frac{1}{3}\sum 1$

$S_{N}=\frac{1}{3}(4+4^{2}+4^{3}+....+4^{N})-\frac{N}{3}$      -(2)

The equation-

$4+4^{2}+4^{3}+....+4^{N}$

is in G.P. with 

First term a = 4

Common ratio r = 4

Applying the formula of sum of G.P.-

$S_{N}=\frac{4(4^{N}-1)}{4-1}$                                               -(3)

Substituting equation (3) in equation (2), we get-

$S_{N}=\frac{1}{3}(\frac{4(4^{N}-1)}{4-1})-\frac{N}{3}$

$S_{N}=\frac{4}{3}(\frac{4^{N}-1}{4-1})-\frac{N}{3}$

$S_{N}=\frac{4}{9}(4^{N}-1)-\frac{N}{3}$

插图：

Input: N = 3
Output: 11
Explanation:
$S_{N}=\frac{4}{9}(4^{N}-1)-\frac{N}{3}$
$S_{N}=\frac{4}{9}(4^{3}-1)-\frac{3}{3}$
$S_{N}=\frac{4}{9}(63)-1$
$S_{N}=27$

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate the sum
// of first N term
int calcSum(int n)
{
    int a = pow(4, n);
    return (4 * (a - 1) - 3 * n) / 9;
}
 
// Driver Code
int main()
{
    // Value of N
    int N = 3;
 
    // Function call to calculate
    // sum of the series
    cout << calcSum(N);
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
 
class GFG{
 
  // Function to calculate the sum
  // of first N term
  static int calcSum(int n)
  {
    int a = (int)Math.pow(4, n);
    return (4 * (a - 1) - 3 * n) / 9;
  }
 
 
  // Driver Code
  public static void main(String[] args)
  {
    // Value of N
    int N = 3;
 
    // Function call to calculate
    // sum of the series
    System.out.print(calcSum(N));
  }
}
 
// This code is contributed by code_hunt.

Python3

# Python 3 program for the above approach
 
# Function to calculate the sum
# of first N term
def calcSum(n):
    a = pow(4, n)
    return (4 * (a - 1) - 3 * n) / 9
 
 
# Driver Code
if __name__ == "__main__":
 
    # Value of N
    N = 3
     
    # Function call to calculate
    # sum of the series
    print(calcSum(N))
 
# This code is contributed by Abhishek Thakur.

C#

// C# code for the above approach
using System;
 
class GFG{
 
  // Function to calculate the sum
  // of first N term
  static int calcSum(int n)
  {
    int a = (int)Math.Pow(4, n);
    return (4 * (a - 1) - 3 * n) / 9;
  }
 
 
  // Driver Code
  public static void Main()
  {
    // Value of N
    int N = 3;
 
    // Function call to calculate
    // sum of the series
    Console.Write(calcSum(N));
  }
}
 
// This code is contributed by gfgking

Javascript

// Javascript program to implement
// the above approach
 
// Function to calculate the sum
// of first N term
function calcSum(n)
{
    let a = Math.pow(4, n)
    return (4 * (a - 1) - 3 * n) / 9
}
 
// Driver Code
 
// Value of N
let N = 3
 
// Function call to calculate
// sum of the series
document.write(calcSum(N))
 
// This code is contributed by samim2000.

输出

27

时间复杂度： O(1)
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