Whakaoti i te (2n+2)(2n+1)-(2n+2)*(2n+1)/n+1

$$( 2 n + 2 ) ( 2 n + 1 ) - \frac { ( 2 n + 2 ) \cdot ( 2 n + 1 ) } { n + 1 }$$

Aromātai

$$2n\left(2n+1\right)$$

Ngā Hipanga Rongoā Poto

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{\left(2n+2\right)\left(2n+1\right)}{n+1}.

$$\left(2n+2\right)\left(2n+1\right)-\frac{2\left(n+1\right)\left(2n+1\right)}{n+1}$$

Me whakakore tahi te n+1 i te taurunga me te tauraro.

$$\left(2n+2\right)\left(2n+1\right)-2\left(2n+1\right)$$

Me whakaroha te kīanga.

$$\left(2n+2\right)\left(2n+1\right)-\left(4n+2\right)$$

Hei kimi i te tauaro o 4n+2, kimihia te tauaro o ia taurangi.

$$\left(2n+2\right)\left(2n+1\right)-4n-2$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 2n+2 ki ia tau o 2n+1.

$$4n^{2}+2n+4n+2-4n-2$$

Pahekotia te 2n me 4n, ka 6n.

$$4n^{2}+6n+2-4n-2$$

Pahekotia te 6n me -4n, ka 2n.

$$4n^{2}+2n+2-2$$

Tangohia te 2 i te 2, ka 0.

$$4n^{2}+2n$$

Whakaroha

$$4n^{2}+2n$$

Ngā Hipanga Rongoā Poto

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{\left(2n+2\right)\left(2n+1\right)}{n+1}.

$$\left(2n+2\right)\left(2n+1\right)-\frac{2\left(n+1\right)\left(2n+1\right)}{n+1}$$

Me whakakore tahi te n+1 i te taurunga me te tauraro.

$$\left(2n+2\right)\left(2n+1\right)-2\left(2n+1\right)$$

Me whakaroha te kīanga.

$$\left(2n+2\right)\left(2n+1\right)-\left(4n+2\right)$$

Hei kimi i te tauaro o 4n+2, kimihia te tauaro o ia taurangi.

$$\left(2n+2\right)\left(2n+1\right)-4n-2$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o 2n+2 ki ia tau o 2n+1.

$$4n^{2}+2n+4n+2-4n-2$$

Pahekotia te 2n me 4n, ka 6n.

$$4n^{2}+6n+2-4n-2$$

Pahekotia te 6n me -4n, ka 2n.

$$4n^{2}+2n+2-2$$

Tangohia te 2 i te 2, ka 0.

$$4n^{2}+2n$$