Đáp án+Giải thích các bước giải:
$a)\\+)1+2+3+\dots+n\\=(n+1).n:2\\=\dfrac{(n+1).n}2$
$+)1+3+5+\dots(2n-1)\\\text{Số số hạng là:}\\(2n-1-1):2+1\\=(2n-2):2+1\\=n-1+1\\=n(số hạng)\\=>(2n-1+1).n:2\\=2n.n:2\\=n^2$
$b)\\+)Đặt, A=1.2+2.3+3.4+\dots+n.(n+1)\\3A=1.2.3+2.3.(4-1)+3.4.(5-2)+\dots+n.(n+1)[(n+2)-(n-1)]\\3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+\dots+n.(n+1).(n+2)-n.(n+1).(n-1)\\3A=n.(n+1).(n+2)\\=>A=\dfrac{n.(n+1).(n+2)}3$
$+)Đặt, B=1.2.3+2.3.4+3.4.5+\dots+n.(n+1).(n+2)\\4B=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+\dots+n.(n+1).(n+2).[(n+3)-(n-1)]\\4B=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+\dots+n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)\\4B=n.(n+1).(n+2).(n+3)\\=>B=\dfrac{n.(n+1).(n+2).(n+3)}4$
