Question

请教一道高中数学关于数列的题目

Sn=1/1*2+1/2*3+1/3*4+…+1/n*(n+1)， n属于正整数。求S10

Answer 1

解：1/[n(n+1)]=(1/n)-[1/(n+1)].如1/(1×2)=1-(1/2).1/(2×3)=(1/2)-(1/3).1/(3×4)=(1/3)-(1/4)....1/[n(n+1)]=(1/n)-[1/(n+1)].将这些等式累加得：Sn=1/1×2+1/2×3+1/3×4+1/4×5+...+1/n×(n+1)=1-[1/(n+1)].当n=10时，S10=1-(1/11)==10/11.

Answer 2

所有这种都可以化简
1/(a*a+t)+1/(a+t)*(a+2t)+1/(a+2t)*(a+3t)
=1/t*(1/a-1/(a+t)+1/(a+t)-1/(a+2t)+1/(a+2t)-1/(a+3t))
=1/t(1/a-1/(a+3t))
如1/3*5=1/2[(5-3)/3*5]=1/2*(5/3*5-3/3*5)=1/2*(1/3-1/5)
题目所求就如一楼回答那样