The answer depends on n. If n is an integer or half on integer then 2n is a whole number; if not, then it isn't.
Oh, dude, the greatest whole number that MUST be a factor of the sum of any six consecutive positive odd numbers is 3. Why? Because any set of consecutive odd numbers will include multiples of 3, ensuring that the sum will be divisible by 3. So, like, 3 is the magic number here.
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
let the number be n. Then: 2n = 3n ⇒ 2n - 2n = 3n - 2n ⇒ 0 = n ie, the number is zero.
The nth hexagonal number is given by the formula: hn = 2n * (2n - 1) / 2
To find the number of whole numbers less than 100 that are 10 greater than an odd whole number, we can set up an equation. Let x represent the odd whole number. Therefore, the number that is 10 greater than x is x + 10. Since x is odd, it can be represented as 2n + 1, where n is a non-negative integer. Substituting this into the equation x + 10 = 2n + 1 + 10, we get 2n + 11. To find the number of such numbers less than 100, we need to find the largest possible value of n such that 2n + 11 < 100. Solving for n, we get n < 44.5, which means the largest possible integer value for n is 44. Therefore, there are 45 odd whole numbers less than 100 that are 10 greater than an odd whole number.
Odd. For any given whole number n, the sum of it and the next whole number is 2n+1. Since 2n is obviously a multiple of 2 and therefore even, 2n+1 must be odd.
Oh, dude, the greatest whole number that MUST be a factor of the sum of any six consecutive positive odd numbers is 3. Why? Because any set of consecutive odd numbers will include multiples of 3, ensuring that the sum will be divisible by 3. So, like, 3 is the magic number here.
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
If the number is n, then twice the number is 2n and the sum of the two is n + 2n = 3n.
let the number be n. Then: 2n = 3n ⇒ 2n - 2n = 3n - 2n ⇒ 0 = n ie, the number is zero.
Every number multiplied by 2 is an even number, so that 2n is an even number. Usually we represent an even number by 2n.
The polygon may have 2n sides, where n is any whole number greater than 1
Any even number can be written in the form 2n for some natural number n.Any odd number can be written as 2n+1 for a natural number nNow add an even to an odd.2n+2n+1=4n+1 which is 2(2n)+1 and this is the form for an odd number.
The nth hexagonal number is given by the formula: hn = 2n * (2n - 1) / 2
To find the number of whole numbers less than 100 that are 10 greater than an odd whole number, we can set up an equation. Let x represent the odd whole number. Therefore, the number that is 10 greater than x is x + 10. Since x is odd, it can be represented as 2n + 1, where n is a non-negative integer. Substituting this into the equation x + 10 = 2n + 1 + 10, we get 2n + 11. To find the number of such numbers less than 100, we need to find the largest possible value of n such that 2n + 11 < 100. Solving for n, we get n < 44.5, which means the largest possible integer value for n is 44. Therefore, there are 45 odd whole numbers less than 100 that are 10 greater than an odd whole number.
0
2n is 2 times n.