just don't do the question.
The sequence 13579 consists of the first five odd numbers. To find the nth term of this sequence, you can use the formula ( a_n = 2n - 1 ), where ( n ) is the term number. Thus, for any positive integer ( n ), the nth term will be the ( n )th odd number. For example, when ( n = 1 ), ( a_1 = 1 ); when ( n = 2 ), ( a_2 = 3 ); and so on.
The 78th odd number can be calculated using the formula 2n-1, where n represents the position of the odd number in the sequence. Therefore, for the 78th odd number, n would be 39 since odd numbers occur at every other position. Plugging 39 into the formula, we get 2(39)-1 = 77. Hence, the 78th odd number is 77.
The 119th odd natural number can be found using the formula for the nth odd number, which is given by (2n - 1). Substituting (n = 119), we get (2 \times 119 - 1 = 238 - 1 = 237). Therefore, the 119th odd natural number is 237.
For any number n you could use * (n % 2 == 0), which would be true for an even number, false for odd For an integer i, a simpler method would be * (i & 1), which would be true for an odd number, false for even
29
1 is the 1st odd number 3 is the 2nd odd number 5 is the 3rd odd number and so on 7 is the 4th odd number and so on if x is the 1000th odd number then x=2*1000-1=1999
The set of odd numbers is an arithmetic sequence. Let say that the sequence has n odd numbers where the first term is a1 and the last one is n. The formula to find the sum on nth terms for an arithmetic sequence is: Sn = (n/2)(a1 + an) or Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference that for odd numbers is 2. Sn = (n/2)(2a1 + 2n - 2)
The 119th odd natural number can be calculated using the formula (2n-1), where n represents the position of the number in the sequence of odd natural numbers. In this case, when n=119, the 119th odd natural number would be (2*119-1) = 237. Therefore, the 119th odd natural number is 237.
The sequence 13579 consists of the first five odd numbers. To find the nth term of this sequence, you can use the formula ( a_n = 2n - 1 ), where ( n ) is the term number. Thus, for any positive integer ( n ), the nth term will be the ( n )th odd number. For example, when ( n = 1 ), ( a_1 = 1 ); when ( n = 2 ), ( a_2 = 3 ); and so on.
printf ("%d is %s\n", n, n%2 ? "odd": "even");
Oh, what a happy little question! To find the 500th odd number, we can use a simple formula: 2n-1, where n is the desired position of the odd number. So for the 500th odd number, we would do 2 * 500 - 1 to get 999. Isn't that just a delightful number to paint with?
There is no such number. Given any odd number, n, the number (n + 2) is a greater odd number. You can go on, for ever, finding larger odd numbers.
The 78th odd number can be calculated using the formula 2n-1, where n represents the position of the odd number in the sequence. Therefore, for the 78th odd number, n would be 39 since odd numbers occur at every other position. Plugging 39 into the formula, we get 2(39)-1 = 77. Hence, the 78th odd number is 77.
#include <stdio.h> int main() { printf("Program to find ODD or Even Number\n"); while(1) { int n = 0; printf("\nEnter a number(-1 for Exit): "); scanf("%d",&n); if( n 0) { printf("%d is a EVEN number.\n", n); } else { printf("%d is a ODD number.\n", n); } } return 0; }
There is a surprisingly easy formula for this. Sum of n odd numbers = n2 So the sum of the first 600 odd numbers (starting with 1 as the very first odd number) is 6002 = 360000.
The 119th odd natural number can be found using the formula for the nth odd number, which is given by (2n - 1). Substituting (n = 119), we get (2 \times 119 - 1 = 238 - 1 = 237). Therefore, the 119th odd natural number is 237.
For any number n you could use * (n % 2 == 0), which would be true for an even number, false for odd For an integer i, a simpler method would be * (i & 1), which would be true for an odd number, false for even