Tính: $\frac{(n-1).n.(n+1)}{4}$ + $\frac{n.(n+1)}{2}$ 27/08/2021 Bởi Rylee Tính: $\frac{(n-1).n.(n+1)}{4}$ + $\frac{n.(n+1)}{2}$
Giải thích các bước giải: $\dfrac{\left ( n – 1 \right ).n.\left ( n + 1 \right )}{4} + \dfrac{n.\left ( n + 1 \right )}{2}$ $= \dfrac{\left ( n – 1 \right ).n.\left ( n + 1 \right )}{4} + \dfrac{2n.\left ( n + 1 \right )}{4}$ $= \dfrac{\left ( n – 1 \right ).n.\left ( n + 1 \right ) + 2n.\left ( n + 1 \right )}{4}$ $= \dfrac{\left ( n^{2} – n \right ).\left ( n + 1 \right ) + 2n.\left ( n + 1 \right )}{4}$ $= \dfrac{\left ( n^{2} – n + 2n \right ).\left ( n + 1 \right )}{4}$ $= \dfrac{\left ( n^{2} + n \right ).\left ( n + 1 \right )}{4}$ $= \dfrac{n.\left ( n + 1 \right ).\left ( n + 1 \right )}{4}$ $= \dfrac{n.\left ( n + 1 \right )^{2}}{4}$ Bình luận
Giải thích các bước giải:
$\dfrac{\left ( n – 1 \right ).n.\left ( n + 1 \right )}{4} + \dfrac{n.\left ( n + 1 \right )}{2}$
$= \dfrac{\left ( n – 1 \right ).n.\left ( n + 1 \right )}{4} + \dfrac{2n.\left ( n + 1 \right )}{4}$
$= \dfrac{\left ( n – 1 \right ).n.\left ( n + 1 \right ) + 2n.\left ( n + 1 \right )}{4}$
$= \dfrac{\left ( n^{2} – n \right ).\left ( n + 1 \right ) + 2n.\left ( n + 1 \right )}{4}$
$= \dfrac{\left ( n^{2} – n + 2n \right ).\left ( n + 1 \right )}{4}$
$= \dfrac{\left ( n^{2} + n \right ).\left ( n + 1 \right )}{4}$
$= \dfrac{n.\left ( n + 1 \right ).\left ( n + 1 \right )}{4}$
$= \dfrac{n.\left ( n + 1 \right )^{2}}{4}$
Giải thích các bước giải: