What is expand ((n+1)(n)b-nc+c)/((n+2)(n+1)) ?

The solution to expand ((n+1)(n)b-nc+c)/((n+2)(n+1)) is (n^2b+nb-nc+c)/(n^2+3n+2)

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expand ((n+1)(n)b-nc+c)/((n+2)(n+1))

expand

\frac{( n + 1 ) ( n ) b - n c + c}{( n + 2 ) ( n + 1 )}

\frac{n ^{2} b + nb - n c + c}{n ^{2} + 3 n + 2}

Solution steps

\frac{( n + 1 ) ( n ) b - n c + c}{( n + 2 ) ( n + 1 )}

Remove parentheses:

(a) = a

= \frac{( n + 1 ) nb - n c + c}{( n + 2 ) ( n + 1 )}

Expand

(n + 2) (n + 1) : n^{2} + 3 n + 2

= \frac{nb ( n + 1 ) - n c + c}{n ^{2} + 3 n + 2}

Expand

(n + 1) nb - n c + c : n^{2} b + nb - n c + c

= \frac{n ^{2} b + nb - n c + c}{n ^{2} + 3 n + 2}

\frac{1}{( s + 1 ) ( s + 0.7 )}

- \frac{2}{3 ( x ^{2} + 4 ) ( x ^{2} + 9 )}

\frac{2.4 + 0.7}{0.5} - \frac{3.6}{1.6}

\frac{d}{d x} ((f (x, y)) (3 x y - 4)^{2})

\frac{1}{( x - 1 ) ( x - 3 ) ( 7 x ^{2} + 6 x + 1 )}

What is expand ((n+1)(n)b-nc+c)/((n+2)(n+1)) ?
The solution to expand ((n+1)(n)b-nc+c)/((n+2)(n+1)) is (n^2b+nb-nc+c)/(n^2+3n+2)