∑k=1nk2=16n(n+1)(2n+1)\displaystyle \sum_{k=1}^n k^2 = \dfrac{1}{6}n(n + 1)(2n + 1)k=1∑nk2=61n(n+1)(2n+1)
∑k=1nk3=[12n(n+1)]2\displaystyle \sum_{k=1}^n k^3 = [\dfrac{1}{2}n(n + 1)]^{2}k=1∑nk3=[21n(n+1)]2
axay=ax+ya^{x} a^{y} = a^{x + y}axay=ax+y
axay=ax−y\dfrac{a^{x}}{a^{y}} = a^{x - y}ayax=ax−y
(ab)x=axbx(ab)^{x} = a^{x}b^{x}(ab)x=axbx
(ax)y=axy(a^{x})^{y} = a^{xy}(ax)y=axy
ab=elnab=eblnaa^{b} = e^{lna^{b}} = e^{blna}ab=elnab=eblna
alogaN=Na^{log_{a}N} = NalogaN=N
logaMn=nlogaMlog_{a}M^{n} = nlog_{a}MlogaMn=nlogaM
loga(MN)=logaM+logaNlog_{a}(MN) = log_{a}M + log_a{N}loga(MN)=logaM+logaN
logaMN=logaM−logaNlog_{a}\dfrac{M}{N} = log_{a}M - log_{a}NlogaNM=logaM−logaN
logab=logcblogcalog_{a}b = \dfrac{log_{c}b}{log_{c}a}logab=logcalogcb（a > 0 且 a ≠ 1; c > 0 且 c ≠ 1; b > 0）
设 a > b > 0，n > 0，则 an>bna^{n} > b^{n}an>bn
设 a > b > 0，n 为正整数，则 an>bn\sqrt[n]{a} > \sqrt[n]{b}na>nb
设 ab<cd\dfrac{a}{b} < \dfrac{c}{d}ba<dc，则 ab<a+cb+d<cd\dfrac{a}{b} < \dfrac{a + c}{b + d} < \dfrac{c}{d}ba<b+da+c<dc
基本不等式及衍生
a+b2≥ab\dfrac{a + b}{2} ≥ \sqrt{ab}2a+b≥ab（其中 a > 0，b > 0；当且仅当 a = b 时等号成立）
a+b+c3≥abc3\dfrac{a + b + c}{3} ≥ \sqrt[3]{abc}3a+b+c≥3abc（其中 a > 0，b > 0，c > 0；当且仅当 a = b = c 时等号成立）
a2+b22≥a+b2≥ab≥21a+1b\sqrt{\dfrac{a^{2} + b^{2}}{2}} ≥ \dfrac{a + b}{2} ≥ \sqrt{ab} ≥ \dfrac{2}{\dfrac{1}{a} + \dfrac{1}{b}}2a2+b2≥2a+b≥ab≥a1+b12（其中 a > 0，b > 0；当且仅当 a = b 时等号成立）
4ab≤(a+b)2≤2(a2+b2)4ab ≤ (a + b)^{2} ≤ 2(a^{2} + b^{2})4ab≤(a+b)2≤2(a2+b2)（当且仅当 a = b 时等号成立）
a2+b2+c2≥13(a+b+c)2≥ab+bc+caa^{2} + b^{2} + c^{2} ≥ \dfrac{1}{3}(a + b + c)^{2} ≥ ab + bc + caa2+b2+c2≥31(a+b+c)2≥ab+bc+ca（当且仅当 a = b = c 时等号成立）
绝对值不等式
−∣a∣≤a≤∣a∣-|a| ≤ a ≤ |a|−∣a∣≤a≤∣a∣
∣a+b∣≤∣a∣+∣b∣|a + b| ≤ |a| + |b|∣a+b∣≤∣a∣+∣b∣
∣a∣−∣b∣≤∣a−b∣≤∣a∣+∣b∣|a| - |b| ≤ |a - b| ≤ |a| + |b|∣a∣−∣b∣≤∣a−b∣≤∣a∣+∣b∣
∣ba+ab∣≥2|\dfrac{b}{a} + \dfrac{a}{b}| ≥ 2∣ab+ba∣≥2（当且仅当 |a| = |b| 时等号成立）
an=a1+(n−1)d=am+(n−m)d=dn+(a1−d)，n∈N∗a_{n} = a_{1} + (n - 1)d = a_{m} + (n - m)d = dn + (a_{1} - d)，n∈N^{*}an=a1+(n−1)d=am+(n−m)d=dn+(a1−d)，n∈N∗
Sn=n(a1+an)2=na1+n(n−1)2d=d2n2−(a1−d2)n，n∈N∗S_{n} = \dfrac{n(a_{1} + a_{n})}{2} = na_{1} + \dfrac{n(n - 1)}{2}d = \dfrac{d}{2}n^{2} - (a_{1} - \dfrac{d}{2})n，n∈N^{*}Sn=2n(a1+an)=na1+2n(n−1)d=2dn2−(a1−2d)n，n∈N∗
an=a1qn−1=amqn−m (a1,q≠0)a_{n} = a_{1}q^{n - 1} = a_{m}q^{n - m} \space (a_{1},q ≠ 0)an=a1qn−1=amqn−m (a1,q=0)
Sn={a1(1−qn)1−q (q≠1),na1 (q=1)S_{n}=\begin{cases} \dfrac{a_{1}(1 - q^{n})}{1-q} \space (q ≠ 1), \\ na_{1} \space (q = 1) \end{cases}Sn=⎩⎨⎧1−qa1(1−qn) (q=1),na1 (q=1)
1n(n+1)=1n−1n+1\dfrac{1}{n(n + 1)} = \dfrac{1}{n} - \dfrac{1}{n + 1}n(n+1)1=n1−n+11
1n(n+k)=1k(1n−1n+k)\dfrac{1}{n(n + k)} = \dfrac{1}{k} (\dfrac{1}{n} - \dfrac{1}{n + k})n(n+k)1=k1(n1−n+k1)
1nm(n+k)=1k(1nm−1m(n+k))\dfrac{1}{nm(n + k)} = \dfrac{1}{k} (\dfrac{1}{nm} - \dfrac{1}{m(n + k)})nm(n+k)1=k1(nm1−m(n+k)1)
1(2n−1)(2n+1)=12(12n−1−12n+1)\dfrac{1}{(2n - 1)(2n + 1)} = \dfrac{1}{2} (\dfrac{1}{2n-1} - \dfrac{1}{2n + 1})(2n−1)(2n+1)1=21(2n−11−2n+11)
1n(n+1)(n+2)=12(1n(n+1)−1(n+1)(n+2))\dfrac{1}{n(n + 1)(n + 2)} = \dfrac{1}{2} (\dfrac{1}{n(n + 1)} - \dfrac{1}{(n + 1)(n + 2)})n(n+1)(n+2)1=21(n(n+1)1−(n+1)(n+2)1)
1a+b=1a−b(a−b)\dfrac{1}{\sqrt{a} + \sqrt{b}} = \dfrac{1}{a - b} (\sqrt{a} - \sqrt{b})a+b1=a−b1(a−b)
1n+n+1=n+1−n\dfrac{1}{\sqrt{n} + \sqrt{n + 1}} = \sqrt{n + 1} - \sqrt{n}n+n+11=n+1−n
1n+n+k=1k(n+k−n)\dfrac{1}{\sqrt{n} + \sqrt{n + k}} = \dfrac{1}{k} (\sqrt{n + k} - \sqrt{n})n+n+k1=k1(n+k−n)
ak=1a−1(ak+1−ak)a^{k} = \dfrac{1}{a - 1} (a^{k + 1} - a^{k})ak=a−11(ak+1−ak)
n⋅n!=(n+1)!−n!n \cdot n! = (n + 1)! - n!n⋅n!=(n+1)!−n!
思路方法 三角函数