Question

求数列前N项的和

（1）1/1*2+1/2*3+1/3*4+...+1/N(N+1)
(2)1/1*3+1/3*5+1/5*7+...+1/(2N-1}(2N+1)

Answer 1

于是调和级数的前n项部分和满足
　　Sn=1+1/2+1/3+…+1/n>ln(1+1)+ln(1+1/2)+ln(1+1/3)+…+ln(1+1/n)
　　=ln2+ln(3/2)+ln(4/3)+…+ln[(n+1)/n]
　　=ln[2*3/2*4/3*…*(n+1)/n]=ln(n+1)
这个是高中常用的不等式，具体和没办法求出把

Answer 2

1/1*2+1/2*3+1/3*4+...+1/N(N+1)
=（1-1/2）+（1/2-1/3）+（1/3-1/4）+.。。+（1/n-1/n+1)
=1-1/n+1
=n/n+1

1/1*3+1/3*5+1/5*7+...+1/(2N-1}(2N+1)
=1/2*[1-1/3+1/3-1/5+.....+1/2n-1-1/2n+1]
=1/2*(1-1/2n+1)
=1/2*2n/2n+1
=n/2n+1

Answer 3

1）1/1*2=1-1/2 1/2*3=1/2-1/3 1/3*4=1/3-1/4则原式=1-1/2+1/2-1/3.....+1/n-1/(n+1)=1-1/(n+1)=n/(n+1)
2）同1）1/3*5=1/2(1/3-1/5)

Answer 4

求数列的前n项和