Question

(ii) (n + 2)! = 60(n-1)!
Find the value of n.
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Answer 1

GIVEN :–

• (n + 2)! = 60(n - 1) !

TO FIND :–

• n = ?

SOLUTION :–

$\\ \implies { \bold{(n + 2)! = 60(n - 1)!}}$

• We know that –

$\\ \implies { \bold{n \: ! = (n)(n - 1)(n - 2).......}}$

• So that –

$\\ \implies { \bold{(n + 2)(n + 1)(n)(n - 1)!= 60(n - 1)!}}$

$\\ \implies { \bold{(n + 2)(n + 1)(n)= 60}}$

• Use hit & trial method –

$\\ \implies \large{ \boxed{ \bold{n=3}}}$

VERIFICATION :–

• Put n = 3 –

$\\ \implies { \bold{(3 + 2)(3 + 1)(3)= 60}}$

$\\ \implies { \bold{(5)(4)(3)= 60}}$

$\\ \implies { \bold{(20)(3)= 60}}$

$\\ \implies { \bold{60= 60 \: \: (verified)}}$

▪︎ Hence, The value of 'n' is 3.

Answer 2

★Answer:★

»We have,

(n+2)!=60(n−1)

As we know n!=(n−1) n!

Therefore,

(n+2)(n+1)n×(n−1)!=60(n−1)!

(n+2)(n+1)n=60

(n²+2n+n+2)n=60

(n³+3n² +2n−60)=0

⇒n³+3n²+2n−60=0

★Factoring the left hand side, this equation resolves to

(n−3)(n²+6n+20)=0(n−3)(n²+6n+20)=0

»Since it can be shown that(n²+6n+20)=0 has (n²+6n+20)=0 no real solution,

n−3=0

n=3

★Hence, the value of n is 3.★