1(1−ax)m=∑nCn+m−1m−1anxn\frac{1}{(1-ax)^m}=\sum_nC_{n+m-1}^{m-1}a^nx^n(1−ax)m1=n∑Cn+m−1m−1anxn
ex=∑i>=0xii!e^x=\sum_{i>=0}\frac{x^i}{i!}ex=i>=0∑i!xi
ex+e−x2=∑i>=0x2i(2i)!\frac{e^x+e^{-x}}{2}=\sum_{i>=0}\frac{x^{2i}}{(2i)!}2ex+e−x=i>=0∑(2i)!x2i
ex−e−x2=∑i>=0x2i+1(2i+1)!\frac{e^x-e^{-x}}{2}=\sum_{i>=0}\frac{x^{2i+1}}{(2i+1)!}2ex−e−x=i>=0∑(2i+1)!x2i+1
ln(x)=∑i>=1(−1)i+1(x−1)ii\ln(x)=\sum_{i>=1}(-1)^{i+1}\frac{(x-1)^i}iln(x)=i>=1∑(−1)i+1i(x−1)i
sin(x)=∑i>=0(−1)ix2i+1(2i+1)!sin(x)=\sum_{i>=0}(-1)^i\frac{x^{2i+1}}{(2i+1)!}sin(x)=i>=0∑(−1)i(2i+1)!x2i+1
cos(x)=∑i>=0(−1)ix2i(2i)!cos(x)=\sum_{i>=0}(-1)^i\frac{x^{2i}}{(2i)!}cos(x)=i>=0∑(−1)i(2i)!x2i
−a1−ax=[ln(1−ax)]′\frac{-a}{1-ax}=\left[ \ln(1-ax)\right]'1−ax−a=[ln(1−ax)]′
(a−x)ex=∑i>=0(a−i)xii!(a-x)e^x=\sum_{i>=0}\frac{(a-i)x^i}{i!}(a−x)ex=i>=0∑i!(a−i)xi
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