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

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
= = = = = = = 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 + ...#includeusing namespace std; // calculate sum upto N term of seriesvoid Sum_upto_nth_Term(int n){    int r = n * (n + 1) *                (2 * n + 7) / 3;    cout << r;} // Driver codeint 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 seriesstatic 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 seriesdef Sum_upto_nth_Term(n):    return n * (n + 1) * (2 * n + 7) // 3 # Driver codeN = 5print(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 seriesstatic void Sum_upto_nth_Term(int n){    int r = n * (n + 1) *                (2 * n + 7) / 3;    Console.Write(r);} // Driver codepublic static void Main(){    int N = 5;    Sum_upto_nth_Term(N);}} // This code is contributed// by Akanksha Rai(Abby_akku)



Javascript


Output:
170

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