Question

1.首项是a1,公差是d的等差数列的通项公式an=a1+(n-1)d,前n项和的公式是Sn=na1+(n(n-1))/2

Answer 1

当n=1时,S1=1*a1+(1*(1-1))/2d=a1

假设当n=k时,Sn=ka1+(k(k-1))/2*d

那么,当n=k+1时,有Sn=(k+1)a1+((k+1)k)/2*d=ka1+(k(k-1))/2*d+a1+(k+1-1)d

∵当n=k时,Sn=ka1+(k(k-1))/2*d

∴,当n=k+1时,Sn=ka1+(k(k-1))/2*d+a1+(k+1-1)d

,有Sn=(k+1)a1+((k+1)k)/2*d

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