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# The best way to get the Sum of the First nth Time period of a Sequence in Java

## The problem#

Your process is to put in writing a operate that returns the sum of the next sequence as much as nth time period(parameter).

``````Sequence: 1 + 1/4 + 1/7 + 1/10 + 1/13 + 1/16 +...
``````

Guidelines:

• You could spherical the reply to 2 decimal locations and return it as String.
• If the given worth is 0 then it ought to return 0.00
• You’ll solely be given Pure Numbers as arguments.

Examples: (Enter –> Output)

``````1 --> 1 --> "1.00"
2 --> 1 + 1/4 --> "1.25"
5 --> 1 + 1/4 + 1/7 + 1/10 + 1/13 --> "1.57"
``````

## The answer in Java code#

Possibility 1:

``````public class NthSeries {
public static String seriesSum(int n) {
double sum = 0.0;
for (int i = 0; i < n; i++)
sum += 1.0 / (1 + 3 * i);
return String.format("%.2f", sum);
}
}
``````

Possibility 2:

``````import java.util.stream.IntStream;
public class NthSeries {
public static String seriesSum(int n) {
return String.format("%.2f", IntStream.vary(0, n).mapToDouble(x -> 1.0 / (3 * x + 1)).sum());
}
}
``````

Possibility 3:

``````import java.util.*;
public class NthSeries {
public static String seriesSum(int n) {
double sum=0.0;
whereas(n>0){
sum+=1.0/(3*n-2);
n--;
}
return String.format("%.2f",sum ).toString();
}
}
``````

## Take a look at instances to validate our answer#

``````import static org.junit.Assert.*;
import java.util.*;
import org.junit.Take a look at;

public class NthSeriesTest {
@Take a look at
public void test1() {
assertEquals("1.57", NthSeries.seriesSum(5));
}
@Take a look at
public void test2() {
assertEquals("1.77", NthSeries.seriesSum(9));
}
@Take a look at
public void test3() {
assertEquals("1.94", NthSeries.seriesSum(15));
}
}
``````
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