sn=1/（1*2）+1/(2*3)+1/（3*4）+……+1/(n(n+1))怎么解？
sn=1/（1*2）+1/(2*3)+1/（3*4）+……+1/(n(n+1))怎么解？
根据:1/n(n+1)=1/n-1/(n+1)
1/1*2+1/2*3+1/3*4+.....1/n(n+1)
=1/1-1/2+1/2-1/3+1/3-1/4+......+1/(n-1)-1/n
+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
所以
sn=1/（1*2）+1/(2*3)+1/（3*4）+……+1/(n(n+1))
=n/(n+1)
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+┉┉+1/[n(n+1)(n+2)]=?怎么做
不知道啥语言，随便写一下把 int n; double sum,tmp; cin >> n; sum=0; for (int i=1;i<=n;i++) { tmp=i*(i+1)*(i+2); sum += 1/tmp; } cout << sum;
-1/1*2- 1/2*3 -1/3*4-`````-1/n*[n+1]
-（1/1*2+1/2*3+...+1/n*(n+1)
=-(1-1/2+1/2-1/3+1/3-1/4...+1/n-1/n+1)
=-(1+0+0+0+...-1/n+1)
=-(1-1/n+1)
=-n/n+1
这种方法叫裂项法
1/(1脳2)+1/(2脳3)+1/(3脳4)+1/n(n+1)=?
=（1/1-1/2）+（1/2-1/3）+（1/3-1/4）……+[1/n-1/(n+1)]
=1/1-1/2+1/2-1/3+1/3-1/4……+1/n - 1/(n+1)
=1-1/(n+1)
求证：1/(1*2*3)+1/(2*3*4)+.+1/(n*(n+1)*(n+2))<1/4
因为1/[n(n+1)(n+2)]
=1/2*{1/[n(n+1)] - 1/[(n+1)(n+2)]}
所以1/(1*2*3)+1/(2*3*4)+......+1/(n*(n+1)*(n+2))
=1/2*{1/(1*2) - 1/(2*3) + 1/(2*3) - 1/(3*4) + …… +1/[n(n+1)] - 1/[(n+1)(n+2)]}
=1/2*{1/2 - 1/[(n+1)(n+2)]}
=1/4 - 1/[2(n+1)(n+2)]显然小于1/4
1×1/2+1/2×1/3+1/3×1/4+.1/n×1/(n+1)怎么做
1×1/2+1/2×1/3+1/3×1/4+....1/n×1/(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)
an=1/[n^1/2+(n+1)^1/2]，Sn=？
列项法去做，分子分母同乘以√（n+1）-√n，an就变成了{√（n+1）-√n}然后求Sn用累加法
a1=√2-√1
a2=√3-√2
……
an=√（n+1）-√n
累加后Sn=√（n+1） -1
lim(n→∞)(1/(1*2)+1/(2*3)+.+1/(n(n+1)))
1/(1*2)+1/(2*3)+...+1/(n(n+1))
=1-1/2+1/2-1/3+……+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+)
=1/(1+1/n)
所以原式=lim(n→∞)1/(1+1/n)
=1/(1+0)
=1
lim[1/(1*2)+1/(2*3)+……1/n(n+1)]=
1/k(k+1) = 1/k - 1/(k+1)
1/(1*2)+1/(2*3)+……1/n(n+1) = 1 - 1/2 + 1/2 - 1/3 + ...... + 1/n - 1/(n+1)
= 1 - 1/(n+1)
原式 = 1
1/(1*2)+1/(2*3)+3/(3*4)+=……+1/(n*(n+1))的极限
1
