数学知识点English裂项与拆分把每个分数拆成两项之差使中间项相消（如 1/(n×(n+1)) = 1/n − 1/(n+1)），求长串分数和。先要掌握小数、分数的四则运算这个知识点的练习题1等差数列{an}\{a_{n}\}{an}的前nnn项和为SnS_{n}Sn，数列{bn}\{b_{n}\}{bn}为等比数列，Sn=n2+nS_{n}\text{=}n^{2}\text{+}nSn=n2+n，b1=a1,b2=a1a22.b_{1}\text{=}a_{1}\text{,}b_{2}\text{=}\dfrac{a_{1}a_{2}}{2}\text{.}b1=a1,b2=2a1a2.…2已知在数列{an}\{a_{n}\}{an}中，a1=1a_{1}=1a1=1，其前nnn项和为sns_{n}sn，且an=2sn22sn−1(n⩾2)(1)a_{n}= \dfrac {2 s_{ n }^{ 2 }}{2s_{n}-1}(n\geqslant 2)(1)an=2sn−12sn2(n⩾2)(1)证明{1sn}\{ \dfrac {1}{s_{n}}\}{sn1}…3下式的值为多少(1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)(1 + \frac{1}{1}) \cdot (1 + \frac{1}{2}) \cdot (1 + \frac{1}{3}) \cdot (1 + \frac{1}{4}) \cdot (1 + \frac{1}{5}) \cdot (1 + \frac{1}{6})(1+11)⋅(1+21)⋅(1+31)⋅(1+41)⋅(1+51)⋅(1+61)…4已知数列{an}\{a_{n}\}{an}满足a1=1a_{1}=1a1=1，且an+1=2an3an+2(n∈N∗).(1)a_{n+1}=\dfrac{2{{a}_{n}}}{3{{a}_{n}}+2}(n\in N^{*}).(1)an+1=3an+22an(n∈N∗).(1)证明：数列{2an}\left\{\dfrac{2}{{{a}_{n}}}\right\}{an2}…