n3 + 2n2 - 15n = 0 : extract n as a factor of all the left hand terms then, n(n2 + 2n - 15) = 0 : the expression n2 + 2n - 15 can be factored as (n + 5)(n - 3) then, n(n + 5)(n - 3) = 0 : This holds true when either n = 0, or n + 5 = 0 or n - 3 = 0 When n + 5 = 0 then n = -5 : When n - 3 = 0 then n = 3. The solutions are n = 0 or n = -5 or n = 3.
The sequence is too short for a definitive answer. One possibility is Un = n2 - 2n + 2 for n = 1, 2, 3, ... another is Un = (n3 - n + 6)/6 or Un = (n3 - 3n2 + 5)/3 There are many more.
2n/5 or (2/5)*n
If x is the smallest odd integer, then x = 2n + 1 for some integer n. Then the next two odd integers are 2n + 3 and 2n + 5 So the question then becomes: 2n+1 + (2n+3) + (2n+5) = 45 or 6n + 9 = 45 to be solved for n and thence the smallest of the three consecutive odd integers.
Combine like terms to give 3n + 2
7+2n
n3 + 2n2 - 15n = 0 : extract n as a factor of all the left hand terms then, n(n2 + 2n - 15) = 0 : the expression n2 + 2n - 15 can be factored as (n + 5)(n - 3) then, n(n + 5)(n - 3) = 0 : This holds true when either n = 0, or n + 5 = 0 or n - 3 = 0 When n + 5 = 0 then n = -5 : When n - 3 = 0 then n = 3. The solutions are n = 0 or n = -5 or n = 3.
The expression m^2n^3/p^3 divided by mp/n^2 can be simplified as (m^2n^3/p^3) / (mp/n^2). This simplifies to (m^3n^5)/(p^3n) = m^3n^4/p^3.
use mathematicl induction to show that (5/4/4n+1)powerox1/2< or equal(1.3.5....(2n+1)/2.4.6.....(2n)< or equal(3/4/2n+1)power of 1/2
The sequence is too short for a definitive answer. One possibility is Un = n2 - 2n + 2 for n = 1, 2, 3, ... another is Un = (n3 - n + 6)/6 or Un = (n3 - 3n2 + 5)/3 There are many more.
The expression 9+5 is equivalent.
2n/5 or (2/5)*n
The expression is 33.
The expression -28x+35 is equivalent to 7y(-4x+5)
If x is the smallest odd integer, then x = 2n + 1 for some integer n. Then the next two odd integers are 2n + 3 and 2n + 5 So the question then becomes: 2n+1 + (2n+3) + (2n+5) = 45 or 6n + 9 = 45 to be solved for n and thence the smallest of the three consecutive odd integers.
Combine like terms to give 3n + 2
6=(1-2n)+5 6-5=(1-2n)+5-5 1=1-2n 1-1=1-1-2n 0=-2n 0/-2=-2n/-2 0=n