110 = 2*5*11
So, if n=9, (n+1)(n+2) = 10*11 = 110
Or, algebraically,
n^2+3n+2 = 110
n^2+3n-108 = 0
(n+12)(n-9) = 0
n = 9,-12
Factorial notation problem. Please help?
Hi, I am stuck on this maths problem. Solve for n: (n+2)!/(n)! =110. I get as far as (n+2)(n+1) = 110. When I multiply it out I get n^2,3n,2 = 110. There is only one example similar to this in my book and it used a quadratic equation to solve it. I feel like that is the next step. I am just unsure of what to do with the 2 I have left over from multiplying out the binomial. I'm really sorry if this doesn't make sense. It's been several years since I had to study mathematics and I am really rusty. Any help is greatly appreciated.
2 answers
Yes I think it does make sense because
n^2+3n+2 = 110
n^2+3n-108 = 0
(n+12)(n-9) = 0
n = 9,-12 makes all of it come together ❤
n^2+3n+2 = 110
n^2+3n-108 = 0
(n+12)(n-9) = 0
n = 9,-12 makes all of it come together ❤
7 answers