Question

设{an}是等差数列,且首项a1>0,公差d>0求证:1/a1a2+1/a2a3+…+1/anan+1=n/a1(a1+nd)

Answer 1

解:
易知
a2-a1=a3-a2=a4-a3=....=a(n+1)-an=d
且an=(a1)+(n-1)d
∴可得:
1/(a1a2)=[(1/a1)-(1/a2)]/d
1/(a2a3)=[(1/a2)-(1/a3)]/d
1/(a3a4)=[(1/a3)-(1/a4)]/d
..................
1/[a(n-1)an]={[1/a(n-1)]-(1/an)]/d
1/[ana(n+1)]={(1/an)-[1/(an+1)]}/d
上式累加,可得:
左边
=[(1/a1)-[1/a(n+1)]}/d
=[a(n+1)-a1]/{dana(n+1)}
=(nd)/{d(an)a(n+1)}
=n/{(a1)[(a1)+nd]}
=右边

Answer 2

1/a1a2+1/a2a3+…+1/anan+1
= [ (a2-a1)/a1a2+(a3-a2)/a2a3+…+(a(n+1)-a(n))/anan+1 ] /d
=[ 1/a1 - 1/a2 + 1/a2 - 1/a3 +...+ 1/an - 1/a(n+1) ] /d
=[ 1/a1 - 1/a(n+1) ] /d
=(a(n+1)-a1)/a1a(n+1)d
=nd / a1a(n+1)d
=n/a1(a1+nd)

Answer 3

由于{an}是等差数列,且首项a1>0,公差d>0
所以
1/a1a2+1/a2a3+…+1/anan+1
=1/d[1/a1-1/a2+1/a2-1/a3+...+1/an-1/a(n+1)]
=1/d[1/a1-1/a(n+1)]
=n/a1(a1+nd)

Answer 4

观察最后一项1/a1(a1+nd)=1/a1(an+1)
分母为第一项的分母为a1a2第二项的分母为a2a3.....第N项的分母为anan+1,
第一项a1与第n项an+1相乘为最终分母,所以将前两项相加,
得2/(a1a3)再与第三项相加消去a3得3/(a1a4)
再与第四项相加消去a4得4/(a1a5)以此类推可得结果
所以不妨用归纳法证明
这是思路,具体操作你还是自己写写吧.

Answer 5

原式=1/d （1/a1-1/a2+1/a3-1/a4+……+1/an-1/an+1）=结果