Question

If (n+2)!=20.n! Find the value of n

Answer 1

We can expand the factorial of a large number into a small number, as:-

n! =n(n-1)!

So,

here

given that

(n+2)! = 20.n!

(n+2)(n+1)n! = 20.n!

cancel out n ! from both side

(n+2) (n+1) = 20

so

${n}^{2} + 3n + 2 = 20 \\ \\ {n}^{2} + 3 n - 18 = 0 \\ \\ {n}^{2} + 6n - 3n - 18 = 0 \\ \\ (n + 6)( n- 3) = 0 \\ \\ \implies \: n = - 6 \: \: \: \: \: or \: \: \: \: \: n = 3 \\ \\ n \ne - ve \\ \\ \therefore \: n = 3$

(n is not negative because factorial of a negative number is not defined .)

Answer 2

given that

(n+2)! = 20.n!

(n+2)(n+1)n! = 20.n!

(n+2) (n+1) = 20

so

${n}^{2} + 3n + 2 = 20 \\ \\ {n}^{2} + 3 n - 18 = 0 \\ \\ {n}^{2} + 6n - 3n - 18 = 0 \\ \\ (n + 6)( n- 3) = 0 \\ \\ \implies \: n = - 6 \: \: \: \: \: or \: \: \: \: \: n = 3 \\ \\ n \ne - ve \\ \\ \therefore \: n = 3$

mrk it as brainliest