a)
$$\sum_{k=0}^n k=n(n+1)/2$$
Ind. Anfang
1=1*2/2
Ind. Annahme
$$\sum_{k=0}^n k =n(n+1)/2$$$$\sum_{k=0}^{n+1} k =n(n+1)/2+n+1=$$$$n(n+1)/2+(n+1)*2/2=$$$$(n+1)(n+2)/2$$Ind. Schluss
b)
$$\sum_{k=0}^n k^2=n*(n+ 1)*(2n+ 1)/6$$
Ind.Anfang
1=1*2*3/6
Ind Annahme
$$\sum_{k=0}^n k^2=$$$$\sum_{k=0}^{n+1} k^2=$$$$n*(n+ 1)*(2n+ 1)/6+(n+1)^2=$$$$n*(n+ 1)*(2n+ 1)/6+(n+1)*6*(n+1)/6=$$$$(n+ 1)*(n*(2n+ 1)+6*(n+1))/6=$$$$(n+ 1)*(2n^2+ 7n+6)/6=$$$$(n+ 1)*(n+2)(2n+3)/6=$$$$(n+ 1)*((n+1)+1)(2(n+1)+1)/6$$Ind. Schluss
c)
$$\sum_{k=0}^n k(k+1) =$$$$\sum_{k=0}^n k^2+k =$$$$\sum_{k=0}^n k^2+\sum_{k=0}^n k=$$$$n*(n+ 1)*(2n+ 1)/6 +n(n+1)/2=$$$$n*(n+ 1)*(2n+ 1)/6 +3n(n+1)/6=$$$$n*(n+ 1)*(2n+ 4)/6=$$$$n*(n+ 1)*(n+ 2)/3$$
d)
$$\sum_{k=0}^n k(k+1) =$$$$n*(n+ 1)*(n+ 2)/3$$
Ind. Anfang
1*2=1*2*3/3
Ind. Annahme
$$\sum_{k=0}^n k(k+1) =$$$$n*(n+ 1)*(n+ 2)/3$$$$\sum_{k=0}^{n+1} k(k+1) =$$$$n*(n+ 1)*(n+ 2)/3+(n+1)*(n+2)=$$$$n*(n+ 1)*(n+ 2)/3+(n+1)*(n+2)*3/3=$$$$(n+1)*(n+2)*(n+3)/3=$$$$(n+1)*((n+1)+1)*((n+1)+2)/3$$
Ind. Schluss