Question

wap in Java to find sum of

1+1/(1+2)+1/(1+2+3)+1/(1+2+3+........n)

Answer 1

Explanation:

If inverse of a sequence follows rule of an A.P i.e, Arithmetic progression, then it is said to be in Harmonic Progression.In general, the terms in a harmonic progression can be denoted as : 1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d) …. 1/(a + nd).

As Nth term of AP is given as ( a + (n – 1)d) .Hence, Nth term of harmonic progression is reciprocal of Nth term of AP, which is : 1/(a + (n – 1)d)

where “a” is the 1st term of AP and “d” is the common difference.

We can use a for loop to find sum.

// C++ program to find sum of series

#include <iostream>

using namespace std;

// Function to return sum of

// 1/1 + 1/2 + 1/3 + ..+ 1/n

class gfg

{

public : double sum(int n)

{

double i, s = 0.0;

for (i = 1; i <= n; i++)

s = s + 1/i;

return s;

}

};

// Driver code

int main()

{

gfg g;

int n = 5;

cout << "Sum is " << g.

Hope it helps

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