Whakaoti i te =1/6n(n+1)(6n^2+14n-5)

$$= \frac { 1 } { 6 } n ( n + 1 ) ( 6 n ^ { 2 } + 14 n - 5 )$$

Aromātai

$$\frac{n\left(n+1\right)\left(6n^{2}+14n-5\right)}{6}$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{1}{6}n ki te n+1.

$$\left(\frac{1}{6}n^{2}+\frac{1}{6}n\right)\left(6n^{2}+14n-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te \frac{1}{6}n^{2}+\frac{1}{6}n ki te 6n^{2}+14n-5 ka whakakotahi i ngā kupu rite.

$$n^{4}+\frac{10}{3}n^{3}+\frac{3}{2}n^{2}-\frac{5}{6}n$$

Whakaroha

$$n^{4}+\frac{10n^{3}}{3}+\frac{3n^{2}}{2}-\frac{5n}{6}$$

Ngā Upane Otinga

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{1}{6}n ki te n+1.

$$\left(\frac{1}{6}n^{2}+\frac{1}{6}n\right)\left(6n^{2}+14n-5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te \frac{1}{6}n^{2}+\frac{1}{6}n ki te 6n^{2}+14n-5 ka whakakotahi i ngā kupu rite.

$$n^{4}+\frac{10}{3}n^{3}+\frac{3}{2}n^{2}-\frac{5}{6}n$$