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Find the sum of first N terms of the series 2×3 + 4×4 + 6×5 + 8×6 + …

  • Difficulty Level : Easy
  • Last Updated : 26 Feb, 2021

Given an integer N. The task is to find the sum upto N terms of the given series:
 

2×3 + 4×4 + 6×5 + 8×6 + … + upto n terms 
 

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Examples: 
 

Input : N = 5
Output : Sum = 170

Input : N = 10
Output : Sum = 990

 



Let the N-th term of the series be tN
t1 = 2 × 3 = (2 × 1)(1 + 2) 
t2 = 4 × 4 = (2 × 2)(2 + 2) 
t3 = 6 × 5 = (2 × 3)(3 + 2) 
t4 = 8 × 6 = (2 × 4)(4 + 2) 



tN = (2 × N)(N + 2)
The sum of n terms of the series, 
 

 Sn = t1 + t2 +... + tn
    =\;\sum_{k=1}^{n} t_{k}
=\;\sum_{k=1}^{n} 2k(k+2)
=\;\sum_{k=1}^{n} (2k^{2}+4k)
=\;\sum_{k=1}^{n}  2k^{2} + \;\sum_{k=1}^{n}  4k
=2\frac{n(n+1)(2n+1)}{6} + 4\frac{n(n+1)}{2}
=n(n+1)[ \frac{2n+1}{3} +2]
=\frac{n(n+1)(2n+7)}{3}

Below is the implementation of above approach: 
 

C++




// C++ program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
#include<iostream>
using namespace std;
 
// calculate sum upto N term of series
void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    cout << r;
}
 
// Driver code
int main()
{
    int N = 5;
    Sum_upto_nth_Term(N) ;
    return 0;
}

Java




// Java program to find sum upto
// N term of the series:
 
import java.io.*;
 
class GFG {
// calculate sum upto N term of series
static void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    System.out.println(r);
}
 
// Driver code
    public static void main (String[] args) {
    int N = 5;
    Sum_upto_nth_Term(N);
    }
}

Python3




# Python program to find sum upto N term of the series:
# 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
# calculate sum upto N term of series
def Sum_upto_nth_Term(n):
    return n * (n + 1) * (2 * n + 7) // 3
 
# Driver code
N = 5
print(Sum_upto_nth_Term(N))

C#




// C# program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
using System;
 
class GFG
{
// calculate sum upto N term of series
static void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    Console.Write(r);
}
 
// Driver code
public static void Main()
{
    int N = 5;
    Sum_upto_nth_Term(N);
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP




<?php
// PHP program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
function Sum_upto_nth_Term($n)
{
    $r = $n * ($n + 1) *
              (2 * $n + 7) / 3;
    echo $r;
}
 
// Driver code
$N = 5;
Sum_upto_nth_Term($N);
 
// This code is contributed
// by Shashank_Sharma
?>

Javascript




<script>
 
// Javascript program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
// calculate sum upto N term of series
function Sum_upto_nth_Term(n)
{
    let r = n * (n + 1) *
                (2 * n + 7) / 3;
    document.write(r);
}
 
// Driver code
 
    let N = 5;
    Sum_upto_nth_Term(N) ;
 
 
// This code is contributed by Mayank Tyagi
 
</script>
Output: 
170

 




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