Đây , mình giải thích cho :ban chung minh cac dang thuc nay duoc khong, minh muon biet tai sao lai nhu vay..
Ta có :
$(1)=1.2 +2.3+3.4+....+n(n+1) =\frac{1.2.(3-0)+2.3.(4-1)+.....+n(n+1)[(n+2)-(n-1)]}{3}$
$=\frac{1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+....+n(n+1)(n+1)-(n-1)n(n+1)}{3} =\frac{n(n+1)(n+2)}{3}$
$(2)=1^2 +2^2+3^2 +...+n^2 =1(2-1)+2(3-1)+3(4-1)+....+n(n+1-1)=1.2-1+2.3+3.4-3+....+n(n+1)-n=1.2+2.3+...+n(n+1) -(1+2+3+..+n)$
$=(1)-\frac{n(n+1)}{2} =n(n+1)(n+2)-\frac{n(n+1)}{2}=DPCM$
$(3)=1^3 +2^3+....+n^3=1^2.(2-1)+2^2.(3-1)+....+n^2.[(n-1)-1]=1^2.2+2^2.3+....n^2.(n+1)-(1^2+2^2+...+n^2)$
$=1.2(1+0)+2.3.(1+1)+...+n(n+1)[1+(n-1)]-(2) =[1.2+2.3+3.4+4.5+...+n(n+1)]+[1.2.3+2.3.4+3.4.5+....+(n-1)n(n+1)]-(2)$
$=(1)-(2)+[1.2.3+2.3.4+3.4.5+....+(n-1)n(n+1)]=(1)-(2)+\frac{1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+....+(n-1)n(n+1)[(n+2)-(n-2)]}{4}$
$=(1)-(2)+\frac{1.2.3.4-0+2.3.4.5-1.2.3.4+....+(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1)}{4}$
$=\frac{n(n+1)(n+2)}{3} -\frac{n(n+1)(2n+1)}{6} +\frac{(n-1)(n(n+1)(n+2)}{4}$
$=\frac{n(n+1)}{2}+\frac{(n-1)(n(n+1)(n+2)}{4}=(\frac{n(n+1)}{2})^2$
$(4)=1.n+2(n-1)+3(n-2)+.....+n[n-(n-1)]=n(1+2+3+...+n)-(1.2+2.3+...+n(n-1)=\frac{n^2.(n+1)}{2}-\frac{n(n-1)(n+1)}{3}$
$=\frac{n^2.(n+1)}{2} -\frac{n(n-1)(n+1)}{3}=\frac{3n^2.(n+1)-2n(n-1)(n+1)}{6} =\frac{n(n+1)[3n-2(n-1)]}{6}=\frac{n(n+1)(n+2)}{6}$