8/n^3(n(n+1)(2n+1)/6) eching

$$\frac { 8 } { n ^ { 3 } } ( \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } )$$

Baholash

$$\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^{2}}$$

Qisqa yechim qadamlari

n ga n+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\frac{8}{n^{3}}\times \frac{\left(n^{2}+n\right)\left(2n+1\right)}{6}$$

n^{2}+n ga 2n+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\frac{8}{n^{3}}\times \frac{2n^{3}+3n^{2}+n}{6}$$

Suratni maxrajga va maxrajini suratga ko‘paytirish orqali \frac{8}{n^{3}} ni \frac{2n^{3}+3n^{2}+n}{6} ga ko‘paytiring.

$$\frac{8\left(2n^{3}+3n^{2}+n\right)}{n^{3}\times 6}$$

Surat va maxrajdagi ikkala 2 ni qisqartiring.

$$\frac{4\left(2n^{3}+3n^{2}+n\right)}{3n^{3}}$$

Hali faktorlanmagan ifodalarni faktorlang.

$$\frac{4n\left(n+1\right)\left(2n+1\right)}{3n^{3}}$$

Surat va maxrajdagi ikkala n ni qisqartiring.

$$\frac{4\left(n+1\right)\left(2n+1\right)}{3n^{2}}$$

Ifodani kengaytiring.

$$\frac{8n^{2}+12n+4}{3n^{2}}$$

Kengaytirish

$$\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^{2}}$$

Qisqa yechim qadamlari

n ga n+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\frac{8}{n^{3}}\times \frac{\left(n^{2}+n\right)\left(2n+1\right)}{6}$$

n^{2}+n ga 2n+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\frac{8}{n^{3}}\times \frac{2n^{3}+3n^{2}+n}{6}$$

Suratni maxrajga va maxrajini suratga ko‘paytirish orqali \frac{8}{n^{3}} ni \frac{2n^{3}+3n^{2}+n}{6} ga ko‘paytiring.

$$\frac{8\left(2n^{3}+3n^{2}+n\right)}{n^{3}\times 6}$$

Surat va maxrajdagi ikkala 2 ni qisqartiring.

$$\frac{4\left(2n^{3}+3n^{2}+n\right)}{3n^{3}}$$

Hali faktorlanmagan ifodalarni faktorlang.

$$\frac{4n\left(n+1\right)\left(2n+1\right)}{3n^{3}}$$

Surat va maxrajdagi ikkala n ni qisqartiring.

$$\frac{4\left(n+1\right)\left(2n+1\right)}{3n^{2}}$$

Ifodani kengaytiring.

$$\frac{8n^{2}+12n+4}{3n^{2}}$$