Question

用归纳法证明等差数列前N项和公式

急用，快!

Answer 1

Sn=n*a1+n(n-1)d/2

当n=1时S1=a1.成立

假设n=k时，有Sk=k*a1+k(k-1)d/2等式成立
则有n=k+1时
S（k+1)=Sk+a(k+1)=Sk+a1+k*d

=k*a1+k(k-1)d/2+a1+k*d
=(k+1)*a1+((k²-k)d+2k*d)/2
=(k+1)*a1+(k²+k)d/2
=(k+1)*a1+(k+1)kd/2

所以有n=k+1时
S(k+1)==(k+1)*a1+(k+1)kd/2成立

问题得证

Answer 2

a1=k
a2=k+d
a3=k+2d
...
an-1=k+(n-2)d
an=k+(n-1)d
把上式相加得
Sn=a1+a2+a3+..+an=nk+d(1+2+..+(n-1)]
=nk+(n-1)n/2d