For any integer n, the numbers 2n + 1 and 2n + 3 are consecutive odd integers.
15
The four numbers will be 2n, 2(n + 1), 2(n + 2), and 2(n + 3), such that: 2 * 2n + 2(n + 3) = 96 6n = 90 n = 15 So the four numbers are 30, 32, 34, and 36
2n x 2n x 2n = 8n^3
I'm assuming you are solving for n.2n + 6 = 7n - 9 {add nine to each side}2n + 15 = 7n { subtract 2n from each side}15 = 5n {divide each side by 5}3 = n
Answer is 3, 5, 7, 9detailsassume the numbers are2n-3, 2n-1, 2n+1, 2n+3 ......................... (1)(2n-3) 2 + (2n-1) 2 + (2n+1) 2 + (2n+3) 2 = 16416n 2 + 20 = 16416n 2 = 144n2 = 9n = 3Substitute in eq 1 we get the answer above
Let the first odd number be 2n-5 (n ≥ 3), then the 6 consecutive odd numbers are: 2n-5, 2n-3, 2n-1, 2n+1, 2n+3, 2n+5 And their sum is: 2n-5 + 2n-3 + 2n-1 + 2n+1 + 2n+3 + 2n+5 = 12n The greatest common factor for all n(≥ 3) of 12n is 12. Thus 12 is the greatest whole number that MUST be a factor of the sum of any six consecutive positive odd numbers.
27
For any integer n, the numbers 2n + 1 and 2n + 3 are consecutive odd integers.
This is because at least one of them must be divisible by 3. The outline proof is as follows: Consider the consecutive odd numbers, 2n+1, 2n+3, 2n+5 ... Suppose 2n+1 is divisible by 3. That is 2n+1 = 3k for some integer k, then 2n+7 = (2n+1)+6 = 3k+6 = 3*(k+2) so that 2n+7 is also divisible by 3. So the run stops with 2n+3, 2n+5. Suppose, instead, that 2n+1 leaves a remainder of 1 when divided by 3. Then 2n+3 and 2n+9 etc are divisible by 3. Again leaving a run of only two consecutive odd primes. Finally, suppose that 2n+1 leaves a remainder of 2 when divided by 3. Then 2n+5, 2n+11 etc are divisible by 3 leaving runs of two consecutive odd primes.
15/6n + 6/6n = 21/6n = 7/2n
15
The four numbers will be 2n, 2(n + 1), 2(n + 2), and 2(n + 3), such that: 2 * 2n + 2(n + 3) = 96 6n = 90 n = 15 So the four numbers are 30, 32, 34, and 36
2n x 2n x 2n = 8n^3
I'm assuming you are solving for n.2n + 6 = 7n - 9 {add nine to each side}2n + 15 = 7n { subtract 2n from each side}15 = 5n {divide each side by 5}3 = n
Let one integer be n, then the other is 2n + 3 and n(2n + 3) = 90; solve this last equation for n: n(2n + 3) = 90 ⇒ 2n2 + 3n - 90 = 0 ⇒ (2n + 15)(n - 6) = 0 ⇒ n = 6 or n = -7.5 As n must be a (positive) integer, the solution n = -7.5 can be ignored, leaving n = 6, giving 2n + 3 = 15. Thus the two positive integers are 6 and 15.
The nth term is 2n+5 and so the next number is 17