yes
(2n - 3)(n - 6)
When n = 4, 3n = 3*4 = 12
There are many possible answers but a relatively straightforward one is Un = 2n2 - n + 3 for n = 1, 2, 3, ...
24.
32. The formula is 2n2 where n is the principal quantum number
2n2 + 15n + 7 = 2n2 + n + 14n + 7 = (2n2 + n) + (14n + 7) = n(2n + 1) + 7(2n + 1) = (2n + 1)(n + 7)
(n - 4)(2n + 3)
8n2+4=12n 8n2-12n+4=0 2n2-3n+1=0 (2n-1)(n-1)=0 2n-1=0 or n-1=0 2n=1 n=1/2 or n=1
n * 2n = 2n2
2n2 / n = 2n
Let lowest number be n. The larger numbers are n + 2 and n + 4. So 4n2 = (n + 2)2 + (n + 4)2 + 12 = n2 + 4n + 4 + n2 + 8n + 16 + 12 = 2n2 + 12n + 32 So 4n2 - 2n2 - 12n - 32 = 0 ie 2n2 - 12n -32 = 0 which factorises as (2n - 16)(n + 2) The positive solution is thus n = 16/2 = 8 and the other numbers are 10 & 12.
yes
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.
-4
The balance is 2N2 +2O2= 4NO
n=1 is the the lowest level there is.