Question

等差数列{an}中,若a10等于0,则有 a1+a2+a3+…+an=a1+a2+…a19-n(n

Answer 1

设公差为d

a2-a1=d

a3-a2=d

.

an-a(n-1)=d

an=(n-1)d+a1

a10=9d+a1

a1=-9d

an=(n-1)d-9d=(n-10)d

a19-n=(19-n-10)d=(9-n)d

a1+a2+.+an=1/2(a1+an)n=1/2n(-9d+(n-10)d)=1/2n(n-19)d

a1+a2+.+a19-n=1/2(a1+a19-n)(19-n)=1/2(-9d+(9-n)d)(19-n)=1/2(-nd)(19-n)=1/2nd(n-19)

所以得证命题

公比为q,q不等于0,

b2/b1=q

bn=b1q^(n-1)

b9=b1q^8=1

要q=1作为特例求解