Question

(n+3)!−(n+2)!=42[(n+1)!+n!] n=?

Answer 1

$(n+2)!(n+3-1)=42n!(n+1+1)$

$(n+2)(n+1).n!.(n+2)=42.n!(n+2)$

$n+2)(n+1)=42$

$n=5$ dir.

Answer 2

$$(n+3)(n+2)(n+1)n!-(n+2)(n+1)n!=42n![(n+1)+1]$$

$$\Rightarrow$$

$$(n+3)(n+2)(n+1)-(n+2)(n+1)=42[(n+1)+1]$$

$$\Rightarrow$$

$$(n+2)(n+1)[(n+3)-1)]=42(n+2)$$

$$\Rightarrow$$

$$(n+2)(n+1)=42$$

$$\Rightarrow$$

$$n=\ldots$$