Tính: $\frac{(n-1).n.(n+1).(n+2)}{4}$ + $\frac{n.(n+1)}{2}$ 27/08/2021 Bởi Eliza Tính: $\frac{(n-1).n.(n+1).(n+2)}{4}$ + $\frac{n.(n+1)}{2}$
Bạn tham khảo : $\dfrac{(n-1). n.(n+1).(n+2)}{4} + \dfrac{ 2n.(n+1)}{4}$ = $\dfrac{ 2n(n+1) + (n-1). n.(n+1).(n+2)}{4}$ = $\dfrac{ 2n(n+1) + (n-1). n.(n^2+2)}{4}$ = $\dfrac{ 2n(n+1) + (n-1). (n^3+2)}{4}$ = $\dfrac{ (2n^2+2n) + (n^4-2)}{4}$ Bình luận
`\frac{(n-1).n.(n+1)(n+2)}{4}+\frac{n.(n+1)}{2}` `=\frac{[(n-1).(n+1)].n.(n+2)}{4}+\frac{2n.(n+1)}{4}` `=\frac{(n^2-1)(n^2+2n)+(2n^2+2n)}{4}` `= \frac{n^4+2n^3-n^2-2n+2n^2+2n}{4}` `= \frac{n^4+2n^3+n^2}{4}` `= \frac{n^2.(n^2+2n+1)}{4}` `= \frac{n^2.(n+1)^2}{4}` `= \frac{[n.(n+1)]^2}{2^2}=\frac{[n.(n+1)]^2}{(-2)^2}` `= (\frac{n^2+n}{2})^2=(\frac{n^2+n}{-2})^2.` Bình luận
Bạn tham khảo :
$\dfrac{(n-1). n.(n+1).(n+2)}{4} + \dfrac{ 2n.(n+1)}{4}$
= $\dfrac{ 2n(n+1) + (n-1). n.(n+1).(n+2)}{4}$
= $\dfrac{ 2n(n+1) + (n-1). n.(n^2+2)}{4}$
= $\dfrac{ 2n(n+1) + (n-1). (n^3+2)}{4}$
= $\dfrac{ (2n^2+2n) + (n^4-2)}{4}$
`\frac{(n-1).n.(n+1)(n+2)}{4}+\frac{n.(n+1)}{2}`
`=\frac{[(n-1).(n+1)].n.(n+2)}{4}+\frac{2n.(n+1)}{4}`
`=\frac{(n^2-1)(n^2+2n)+(2n^2+2n)}{4}`
`= \frac{n^4+2n^3-n^2-2n+2n^2+2n}{4}`
`= \frac{n^4+2n^3+n^2}{4}`
`= \frac{n^2.(n^2+2n+1)}{4}`
`= \frac{n^2.(n+1)^2}{4}`
`= \frac{[n.(n+1)]^2}{2^2}=\frac{[n.(n+1)]^2}{(-2)^2}`
`= (\frac{n^2+n}{2})^2=(\frac{n^2+n}{-2})^2.`