What is expand ((-1)^{n+2}(n+1)(n+2))/2 ?

The solution to expand ((-1)^{n+2}(n+1)(n+2))/2 is ((-1)^{n+2}n^2)/2+(3(-1)^{n+2}n)/2+(-1)^{n+2}

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expand ((-1)^{n+2}(n+1)(n+2))/2

expand

\frac{( - 1 ) ^{n + 2} ( n + 1 ) ( n + 2 )}{2}

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

Solution steps

\frac{( - 1 ) ^{n + 2} ( n + 1 ) ( n + 2 )}{2}

Expand

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

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

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

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

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

Divide the numbers:

\frac{2}{2} = 1

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

\frac{x ( x ^{2} - 2 )}{y ^{2}}

\frac{y ( x - π )}{x ( 1 - p ^{2} )}

\frac{( 6 + b ) ^{2}}{2 t - 6}

so l v e (lo g_{10} (N) = 0.3 + 2 t, N)

\frac{3 s}{8 ( s + 1 )}

What is expand ((-1)^{n+2}(n+1)(n+2))/2 ?
The solution to expand ((-1)^{n+2}(n+1)(n+2))/2 is ((-1)^{n+2}n^2)/2+(3(-1)^{n+2}n)/2+(-1)^{n+2}