Question

等差数列公式推导过程…急!

Answer 1

a(n)=a+(n-1)d,s(n)=na+n(n-1)d/2.

证明.归纳法.

n=1时,s(1)=a(1)=a,结论成立.

设n=k时结论成立,则s(k)=ka+k(k-1)d/2.

n=k+1时,

s(k+1)=s(k)+a(k+1)=ka+k(k-1)d/2 + a+kd = (k+1)a + k(k-1)d/2 + 2kd/2 = (k+1)a + k[k-1+2]d/2=(k+1)a + k(k+1)d/2,结论成立.

因此,由归纳法知,a(n)=a+(n-1)d.s(n)=na+n(n-1)d/2.