Question

Write a program to find the sum of the series:

s= 1- a + a2

- a3+ ---------------+a10 in java

. write accurately I have to type​

Answer 1

Answer:

important java. util. *

public class series

{

public static void main(String args[])

{

Scanner in=new Scanner(System.in) ;

inter a, s, i;

System.out.println("Enter a no:") ;

a=in.nextInt() ;

for(i=1;i<=10;i++)

{

if(i%2! =0)

{

s=s-(a*i) ;

System.out.println(s) ;

}

else

{

s=s+(a*i) ;

System.out.println(s) ;

}

}

}

}

Explanation:

mark me brainliest for my hardwork

Answer 2

Correct Series:-

$\tt \: s = 1 -a + {a}^{2} - {a}^{3} + ... + {a}^{10}$

Required Answer:-

import java.util.*;

class Series{

public static void main (String ar []){

Scanner sc=new Scanner (System.in);

System.out.println("Value of a?");

int a=sc.nextInt();

double s=0;

for(int I=0;I<=10;I++)

s=(i%2==0)?(s+(Math.pow(a,I))):(s-(Math.pow(a,I)));

System.out.println("Sum="+s);

}

}

Note:-

• You can also write

if (i%2==0)

s+=Math.pow(a,I));

else

s-=Math.pow(a,I))

Logic:-

• Initialise s as 0 and accept value of a as input.
• Run a loop from 0 to 10
• Obviously, any number raised to power 0 will be 1 so at first iteration s will get 1
• Second iteration, I becomes 1 and 1%2==0 condition false so control will execute else part
• Next iteration, I becomes 2 and 2 %2==0, condition true, control executes the if part.
• After loop gets over, display sum
• Woohoo! You made the program!!!

•••♪