Question

if (n+2)!=210(n-1)!,then the value if n is equal to

Answer 1

Here (n+2)! = (n+2)(n+1)n!
And n! = n(n-1)!

So (n+2)! = (n+2)(n+1)n(n-1)!

Resolving the given equation,

(n+2)(n+1)n(n-1)! = 210(n-1)!
(n+2)(n+1)n = 210
(n^2 + 3n+2)n = 210
n^3 + 3n^2 + 2n = 210
n^3 + 3n^2 + 2n - 210 = 0
By trial method, we get n=5.

Hope it helps!!

Answer 2

Answer:

The value of n is 5

Step-by-step explanation:

Given that (n + 2)! = 210 (n - 1)!

=> (n + 2) (n + 1) n (n - 1)! = 210 (n - 1)!

=> (n + 2) (n + 1) n = 210                                          --(i)

=> Product of three consecutive numbers = 210

We know that 7 x 6 x 5 = 210                              --(ii)

Therefore, on comparing equation (i) and (ii), we get,

n + 2 = 7 and n + 1 = 6 and n = 5

Therefore, n = 5

Thus, the value of n is 5