高一数学数列＆框图

Question

请计算第二问，写出步骤，谢谢

Answer 1

这个题目挺好以程序流程图的形式表述实际上an是个等差数列
a1=1；a2=1+d；a3=1+2d；a4=1+3d；a5=1+4
d；没计算一次有个m值；S最后的结果就是将每次的m值和数列的前五项和相加，题目中已给出结果可以解除d，所以同乡也就出来。慢慢读胴体木意思就好做了~~

Answer 2

(2)S=1/(a1*a2)+1/(a2*a3)+1/(a3*a4)+…+1/[ak*a(k+1)]
a2=a1+d,an=a1+(n-1)d
ak=a1+(k-1)d
1/[an*a(n+1)]=1/[an*(an+d)]=(1/d)[1/an-1/(an+d)]
a1=1,k=5
所以S=1/(a1*a2)+1/(a2*a3)+1/(a3*a4)+1/(a4*a5)+1/(a5*a6)
=(1/d)*(1/a1-1/a2+1/a2-1/a3+1/a3-1/a4+1/a4-1/a5+1/a5-1/a6)
=(1/d)*(1/a1-1/a6)
=(1/d)*[1/a1-1/(a1+5d)]
=(1/d)*[1-1/(1+5d)]
=(1/d)*[5d/(1+5d)]
=5/(1+5d)
=5/11
所以1+5d=11
d=2
所以an=a1+(n-1)d=1+2(n-1)=2n-1追问

1/[an*a(n+1)]=1/[an*(an+d)]=(1/d)[1/an-1/(an+d)]
请问这一步是什么意思？

追答

a(n+1)=an+d没问题吧
1/[an*a(n+1)]
=1/[an*(an+d)]
=(1/d)*d*{1/[an*(an+d)]}
=(1/d)*{ d/[an*(an+d)] }
=(1/d)*{ [(an+d)-an] / [an*(an+d)] }
=(1/d)*{ (an+d)/ [an*(an+d)] - an/ [an*(an+d)] }
=(1/d)*[1/an-1/(an+d)]
明白了吧
其实就像1/（2x5)=(1/3)x(1/2-1/5)一样，体会一下。

本回答被提问者采纳