等差数列{1\an}满足a1=1,公差d=2,求a1a2+a2a3+……+anan+1的和
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![等差数列{1\an}满足a1=1,公差d=2,求a1a2+a2a3+……+anan+1的和](/uploads/image/z/1079676-36-6.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BD%9B1%5Can%EF%BD%9D%E6%BB%A1%E8%B6%B3a1%3D1%2C%E5%85%AC%E5%B7%AEd%3D2%2C%E6%B1%82a1a2%2Ba2a3%2B%E2%80%A6%E2%80%A6%2Banan%2B1%E7%9A%84%E5%92%8C)
等差数列{1\an}满足a1=1,公差d=2,求a1a2+a2a3+……+anan+1的和
等差数列{1\an}满足a1=1,公差d=2,求a1a2+a2a3+……+anan+1的和
等差数列{1\an}满足a1=1,公差d=2,求a1a2+a2a3+……+anan+1的和
1/a1=1
d=2
所以1/an=(2n-1)
所以原式=1/1*3+1/3*5+……+1/(2n-1)(2n+1)
=(1/2)(1-1/3)+(1/2)(1/3-1/5)+……+(1/2)[1/(2n-1)-1/(2n+1)]
中间正负抵消
=(1/2)[1-1/(2n+1)]
=n/(2n+1)