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Recently Referred Formulae

a_{n+2}-\alpha a_{n+1}=\alpha(a_{n+1}-\alpha a_n)=\alpha^{n+1}(a_1-\alpha a_0)
a_{n+1}-\frac{1}{2}=-(a_n-\frac{1}{2})=(-1)^{n}(a_1-\frac{1}{2})=\frac{(-1)^{n}}{2}
a_n=a_0p^n+\alpha(1-p^n)=a_0p^n+q\sum_{k=0}^{n-1}p^k;\ \ n=1,\ 1-p=0
\sum_{k=1}^{n-1}(loga_{k+1}-loga_k)=\sum_{k=1}^{n-1}log2=log2\sum_{k=1}^{n-1}1=(n-1)log2=loga_n-loga_1
a_{n+1}/a_n=2