Question

Consider the following sequence:1st term: 12nd term: 1 2 13rd term: 1 2 1 3 1 2 14th term: 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1And so on. Complete the given method solve that takes as parameter an integer n and prints the nth terms of this sequence.Hint: Perhaps you should use a String to store the sequence? What is happening in each term of the sequence?Example Input: 1Output: 1Example Input: 4Output: 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1​

Answer 1

Answer:

The series 3+6+9+12+⋯+30 can be expressed as sigma notation ∑n=1103n . This expression is read as the sum of 3n as n goes from 1 to 10

Example 1:

Find the sum of the first 20 terms of the arithmetic series if a1=5 and a20=62 .

S20=20(5 + 62)2S20=670

Example 2:

Find the sum of the first 40 terms of the arithmetic sequence

2,5,8,11,14,⋯

First find the 40 th term:

a40=a1+(n−1)d        =2+39(3)=119

Then find the sum:

Sn=n(a1+an)2S40=40(2 + 119)2=2420

Example 3:

Find the sum:

∑k=150(3k+2)

First find a1 and a50 :

a1=3(1)+2=5a20=3(50)+2=152

Then find the sum:

Sk=k(a1 + ak)2S50=50(5 + 152)2=3925

Explanation:

Answer 2

Answer:

Here's an example of how the method solve(int n) could be implemented in Java:

Explanation:

public static void solve(int n) {

   StringBuilder sb = new StringBuilder();

   for (int i = 1; i <= n; i++) {

       for (int j = 1; j <= i; j++) {

           sb.append(j).append(" ");

       }

       for (int j = i - 1; j >= 1; j--) {

           sb.append(j).append(" ");

       }

   }

   System.out.println(sb.toString());

}

In this code, we use a StringBuilder to store the sequence. In each iteration of the outer loop, we first append all numbers from 1 to i, then we append all numbers from i-1 to 1.

For example, when n = 4, the output will be "1 2 1 3 1 2 1 4 1 2 1 3 1 2 1"

It is worth to mention that StringBuilder is more efficient than String when it comes to building strings incrementally as it does not need to create a new object on each concatenation.

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