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
=
=
=
=
=
=![n(n+1)[ \frac{2n+1}{3} +2]](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20210%2033'%3E%3C/svg%3E "Rendered by QuickLaTeX.com")
=
=
=
=
=
=![n(n+1)[ \frac{2n+1}{3} +2]](https://neveropen.co.uk/wp-content/uploads/2023/11/wardslaus_quicklatex.com-99c6f9cfaab5d8ed8fc1d09d69e72104_l3.png "Rendered by QuickLaTeX.com")
=
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 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 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
Time Complexity: O(1), it is a constant.
Auxiliary Space: O(1), no extra space is required.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
