Question

Simplify
n²-9/(n+3)! + 6/(n+2)! - 1/(n+1)​

Answer 1

${ \huge{ \boxed{ \bf{\underline{ \red{Answer}}}}}} : -$

Expand (n+3)!  and (n+2)! into the product of natural numbers up to (n+1)!  and (n+1)! 

respectively using tip 1

$\therefore \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}=\frac{{{n}^{2}}-9}{(n+3)(n+2)(n+1)!}+\frac{6}{(n+2)(n+1)!}-\frac{1}{(n+1)!}$

Solve the fundamental operations of RHS from the above equation

$\begin{gathered} \therefore \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}=\frac{{{n}^{2}}-9+6(n+3)-(n+3)(n+2)}{(n+3)(n+2)(n+1)!} \\ \therefore \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}=\frac{{{n}^{2}}-9+6n+18-{{n}^{2}}-5n-6}{(n+3)(n+2)(n+1)!} \\ \therefore \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}=\frac{n+18-15}{(n+3)(n+2)(n+1)!} \\ \therefore \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}=\frac{n+3}{(n+3)(n+2)(n+1)!} \end{gathered}$

$\therefore \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}=\frac{1}{(n+2)(n+1)!}$

Convert (n+2)(n+1)! into (n+1)! using tip 1

$\therefore \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}=\frac{1}{(n+2)!}$

$Hence, the value of expression </strong></p><p><strong>[tex]Hence, the value of expression \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!}</strong></p><p><strong>[tex]Hence, the value of expression \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!} is  </strong></p><p><strong>[tex]Hence, the value of expression \frac{{{n}^{2}}-9}{(n+3)!}+\frac{6}{(n+2)!}-\frac{1}{(n+1)!} is  \frac{1}{(n+2)!}$

Answer 2

Answer:

1/(n+2)!

Step-by-step explanation:

See picture

Mark as brainliest