Question

关于一道数列问题

求数列1/3,1/8,1/15,1/24,1/35......的前N项和Sn.

Answer 1

1/3 + 1/8 + 1/15 + 1/24 + 1/35 + ...
= 1/(1·3) + 1/(2·4) + 1/(3·5) + 1/(4·6) + 1/(5·7) + ... + 1/[n·(n+2)]
= 1/2·【 (3-1)/(1·3) + (4-2)/(2·4) + (5-3)/(3·5) + (6-4)/(4·6) + (7-5)/(5·7) + ... + [(n+2) - n]/[n·(n+2)]】
= 1/2·【1/1 - 1/3 + 1/2 - 1/4 +1/3 - 1/5 + 1/4 - 1/6 + 1/5 - 1/7 + ... + 1/n - 1/(n+2)】
= 1/2·【1+1/2 - 1/(n+1) - 1/(n+2)】
= (3n²+5n)/[4·(n+1)(n+2)]
= (3n²+5n)/(4n²+12n+8)

这是数列求和的裂项法，特点是分母为等差数列的积，关键是把分子写成分母因数的差，然后配相应的系数，最后注意抵消的顺序。

Answer 2

易看出数列的通项公式为
An=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]

∴Sn=A1+A2+...+An
=(1/2){(1-1/3)+(1/2-1/4)+(1/3-1/5)+...+[1/(n-2)-1/n]+[1/(n-1)-1/(n+1)]+[1/n-1/(n+2)]
=(1/2){1+1/2+(-1/3+1/3)+(-1/4+1/4)+...+[-1/(n-1)+1/(n-1)]+(-1/n+1/n)-1/(n+1)-1/(n+2)}
=(1/2)[3/2-1/(n+1)-1/(n+2)]
=(3n^2+5n)/4(n^2+3n+2)

Answer 3

数列1/3,1/8,1/15,1/24,1/35......的通项公式an=1/[(n+1)²-1]
即an=1/n(n+2)=[1/n-1/(n+2)]/2
Sn=a1+a2+...+an
=[1-1/3+1/2-1/4+1/3-1/5...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2
=[3/2-1/(n+1)-1/(n+2)]/2
={[3(n+1)(n+2)-2(n+2)-2(n+1)]/2(n+1)(n+2)}/2
=(3n²+5n)/4(n²+3n+2)
=(3n²+5n)/(4n²+12n+8)

Answer 4

受教

Answer 5

易知数列的通项为1/n(n+2)，裂项得1/2(1/n-1/(n+2))，前N项和为1/2(1-1/3+1/2-1/4+1/3-1/5+……+1/(n-1)-1/(n+1)+1/n-1/(n+2))，中间各项可消去，最后得1/2(3/2-1/(n+1)-1/(n+2))。