0.999...=0.\dot{9}
\frac{1}{3}\quad\quad\quad=0.\dot{3}
\frac{1}{3} * 3=0.\dot{3} * 3
1\quad\quad\quad=0.\dot{9}
1-0.\dot{9}=0.\dot{0}
0.\dot{0} = 0
0.\dot{9}
a_1=0.9
r=0.1
a_1r^{n-1}
S_n=\frac{a_1(r^n-1)}{r-1}
S_n=\frac{a_1(r^n-1)}{r-1}=\frac{0.9(0.1^n-1)}{-0.9}
n
0.\dot{9}=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{0.9(0.1^n-1)}{-0.9}=\lim_{n\to\infty}-(0.1^n-1)=\lim_{n\to\infty}1-0.1^n=1
0.\dot{9}
p\(1\)=\frac{99}{100}
p\(2\)=\frac{1}{100}
p\(3\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}
p\(3\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}=\frac{1}{100}\(1+\frac{1}{99}\)=\frac{1}{100}\(\frac{99}{99}+\frac{1}{99}\)=\frac{1}{100}\cdot\frac{100}{99}=\frac{1}{99}
p\(4\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}+\frac{1}{100}\cdot\frac{1}{98}+\frac{1}{100}\cdot\frac{1}{99}\cdot\frac{1}{98}
p\(3\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}
=\frac{1}{99}
p\(4\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}+\frac{1}{100}\cdot\frac{1}{98}+\frac{1}{100}\cdot\frac{1}{99}\cdot\frac{1}{98}=\frac{1}{99}+\frac{1}{99}\cdot\frac{1}{98}
=\frac{1}{99}\(1+\frac{1}{98}\)=\frac{1}{99}\cdot\frac{99}{98}=\frac{1}{98}
p\(n\)=\frac{1}{102-n}
p\(n\)=p\(n-1\)+p\(n-1\)\cdot\frac{1}{102-n}
p\(n\)=\frac{1}{102-n}
p\(n\)=p\(n-1\)+p\(n-1\)\cdot\frac{1}{102-n}
p\(2\)=\frac{1}{100}
p\(3\)=p\(2\)+p\(2\)\cdot\frac{1}{102-3}=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}=\frac{1}{99}=\frac{1}{102-n}
p\(n+1\)=p\(n\)+p\(n\)\cdot\frac{1}{102-\(n+1\)}=\frac{1}{102-n}+\frac{1}{102-n}\cdot\frac{1}{102-\(n+1\)}=\frac{1}{102-n}\(1+\frac{1}{102-n-1}\)
=\frac{1}{102-n}\(\frac{102-n-1}{102-n-1}+\frac{1}{102-n-1}\)=\frac{1}{102-n}\(\frac{102-n}{102-n-1}\)=\frac{1}{102-\(n+1\)}
p\(100\)=\frac{1}{102-100}=\frac{1}{2}
p\(n\)
p\(1\)=1
p\(2\)=\frac{365}{365}\cdot\frac{364}{365}
p\(3\)=\frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}
p\(n\)=\prod_{i=1}^{n}\frac{366-i}{365}=\frac{1}{365^n}\cdot\frac{365!}{(365-n)!}
1-\frac{1}{365^n}\cdot\frac{365!}{(365-n)!}
\frac{1}{4}
p\(n\)
f\(n\)
1-p=\(1-w^n\)^k
k = \frac{\log(1-p)}{\log(1-w^n)}
\frac{\partial \bf{P}}{\partial \bf{t}}
-\frac{b \pm \sqrt{b^2-4ac}}{2a}
e^{i \pi} + 1 = \cos (\pi) + i\sin(\pi) + 1 = 0
D=(\bf{x} - \bf{\mu})^\top \Sigma^{-1} (\bf{x} - \bf{\mu})
E=\sum_{k=1}^{n+m}p\(k\)k
E=1 \times \frac{1}{2}+2\times \frac{1}{2}=1.5
E=\sum_{k=1}^{m+1}k\frac{1}{m+1}=\frac{1}{m+1}\cdot \frac{1}{2}(m+1)(m+2)=\frac{1}{2}(m+2)
=\frac{1}{2}m+1
E=\sum_{k=2}^{m+2}p\(k\)k=\frac{1}{10}2+\frac{2}{10}3+\frac{3}{10}4+\frac{4}{10}5=\frac{2+6+12+20}{10}=4
E=\frac{1}{10}2+\frac{2}{10}3+\frac{3}{10}4+\frac{4}{10}5=\frac{1}{10}(1\cdot 2 + 2\cdot 3 + 3\cdot 4+4\cdot 5)=4
M=\frac{(m+1)(m+2)}{2}
k\cdot (k-1)
E=\frac{1}{M}\sum_{k=2}^{m+2}k\cdot(k-1)=\frac{2}{(m+1)(m+2)}\sum_{k=2}^{m+2}k\cdot(k-1)
E=\frac{1}{M}\sum_{k=2}^{m+2}k\cdot(k-1)\\=\frac{2}{(m+1)(m+2)}\sum_{k=2}^{m+2}k^2-k=\frac{2}{(m+1)(m+2)}(\sum_{k=2}^{m+2}k^2-\sum_{k=2}^{m+2}k)\\=\frac{2}{(m+1)(m+2)}((\sum_{k=1}^{m+2}k^2-\sum_{k=1}^{1}k^2)-(\sum_{k=1}^{m+2}k-\sum_{k=1}^{1}k))
\Sigma
E=\frac{2}{(m+1)(m+2)}((\frac{1}{6}(m+2)(m+3)(2m+5)-1)-(\frac{1}{2}(m+2)(m+3)-1))\\=\frac{2}{(m+1)(m+2)}(\frac{1}{6}(m+2)(m+3)(2m+5)-\frac{1}{2}(m+2)(m+3))
=\frac{1}{3(m+1)(m+2)}((m+2)(m+3)(2m+5)-3(m+2)(m+3))\\=\frac{(m+2)(m+3)(2m+2)}{3(m+1)(m+2)}=\frac{2(m+2)(m+3)(m+1)}{3(m+1)(m+2)}=\frac{2(m+3)}{3}\\E=\frac{2(m+3)}{3}=\frac{2m}{3}+2
E=\frac{2m}{3}+2=\frac{2\cdot 3}{3}+2=2+2=4
n=1
\frac{1}{2}m+1
n=2
\frac{2}{3}m+2
E=\frac{n}{n+1}m + n
E=\sum_{k=n}^{n+m}p\(k\)k
[n,n+m]
p(k)
p(k)
{}_{n+m} \mathrm{C}_{n}
{}_{a+b} \mathrm{C}_{a}
{}_{n+m+1} \mathrm{C}_{n+1}
{}_{n+m-1}\mathrm{C}_{n-1}
{}_{n+m}\mathrm{C}_{n}
p(k)
p(k)=\frac{{}_{k-1}\mathrm{C}_{n-1}}{{}_{n+m}\mathrm{C}_{n}}
E=\sum_{k=n}^{n+m}p\(k\)k\\=\sum_{k=n}^{n+m}\frac{{}_{k-1}\mathrm{C}_{n-1}}{{}_{n+m}\mathrm{C}_{n}}\cdot k
{}_a \mathrm{C}_b = \frac{a!}{(a-b)!b!}
E=\sum_{k=n}^{n+m}\frac{{}_{k-1}\mathrm{C}_{n-1}}{{}_{n+m}\mathrm{C}_{n}}\cdot k\\=\sum_{k=n}^{n+m}\frac{m!n!}{(n+m)!}\frac{(k-1)!}{(n-1)!(k-n)!}\cdot k\\=\frac{m!n!}{(n+m)!}\sum_{k=n}^{n+m}\frac{(k-1)!}{(n-1)!(k-n)!}\cdot k\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}\frac{(k-1)!}{(k-n)!}\cdot k
\Sigma
(k-1)!k=k!
\frac{k!}{(k-n)!}={}_k \mathrm{P}_n
E=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}\frac{(k-1)!}{(k-n)!}\cdot k\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}\frac{k!}{(k-n)!}\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}{}_k\mathrm{P}_n
\sum_{k=n}^{n+m}{}_k\mathrm{P}_n
\sum_{k=1}^{a}{}_k \mathrm{P}_b=\frac{1}{(b+1)}\overbrace{(a+1)a(a-1)\cdots(a+1-b)}^{b+1}
E=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}{}_k\mathrm{P}_n\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}(\sum_{k=1}^{n+m}{}_k\mathrm{P}_n-\sum_{k=1}^{n-1}{}_k\mathrm{P}_n)\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}(\frac{1}{(n+1)}(n+m+1)(n+m)\cdots(m+1)-\frac{1}{(n+1)}n(n-1)\cdots(n-n))\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\frac{1}{(n+1)}\frac{(n+m+1)!}{m!}\\=\frac{m!n!}{(n-1)!m!}\frac{(n+m+1)!}{(n+m)!}\frac{1}{(n+1)}\\=\frac{n(n+m+1)}{(n+1)}=\frac{nm}{(n+1)}+\frac{n(n+1)}{(n+1)}=\frac{n}{n+1}m+n
\frac{1}{(n+1)}n(n-1)\cdots(n-n)
(n-n)=0
E=\frac{n}{n+1}m + n
\sum_{k=1}^{a}{}_k \mathrm{P}_b=\frac{1}{(b+1)}\overbrace{(a+1)a(a-1)\cdots(a+1-b)}^{b+1}
E=\frac{10}{11}m + 10
p
n
(1+n)^p -n^p -1
p
(1+n)^p -n^p -1
= ({}_pC_01^p + {}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1} + {}_pC_pn^p) \quad - n^p \quad - 1
(1+n)^p -n^p -1
= (1 + {}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1} + n^p) \quad - n^p \quad - 1
= {}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1}
{}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1}
=\sum^{p-1}_{k=1}\left({}_pC_k1^{p-k}n^{k}\right)
=\sum^{p-1}_{k=1}\left({}_pC_kn^{k}\right)
{}_pC_kn^{k}=pm
m
p
{}_pC_kn^{k}
{}_pC_kn^{k}
=\frac{p!}{(p-k)!}\frac{1}{k!}n^k
k \ge 1
p
k \ge 1
p
{}_pC_kn^{k}
=pm
m
(1+n)^p -n^p -1
\begin{array}{crlllllll}p=1:&1&+n&&&&&-n&-1\\p=2:&1&+2n&+n^2&&&&-n^2&-1\\p=3:&1&+3n&+3n^2&+n^3&&&-n^3&-1\\p=4:&1&+4n&+6n^2&+4n^3&+n^4&&-n^4&-1\\p=5:&1&+5n&+10n^2&+10n^3&+5n^4&+n^5&-n^5&-1\\\end{array}
p=4
=4
{}_pC_k
=\frac{p!}{(p-k)!}\frac{1}{k!}
k
0\le k \le
p
1\le k \le p-1
(1+n)^p -n^p -1
=pm
(1+n
p
f(x)=\frac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}
f(1)+f(2)+f(3)+\cdots +f(60
a_n=\sqrt{2n-1}
a_{n+1}\cdot a_n=\sqrt{2n+1}\sqrt{2n-1}=\sqrt{4n^2-1}
a_{n+1}^2+a_n^2=2n+1+2n-1=4n
a_{n+1}^2-a_n^2=2n+1-2n-1=2
f(n
(a_{n+1}-a_n)
f(n)=\frac{(a_{n+1}-a_n)(a_{n+1}^2+a_n^2+a_{n+1}\cdot a_n)}{(a_{n+1}-a_n)(a_{n+1}+a_n)}=\frac{a_{n+1}^3+a_{n+1}^2a_n+a_{n+1}a_n^2-a_{n+1}^2a_n-a_{n+1}a_n^2-a_n^3}{a_{n+1}^2-a_n^2}
=\frac{a_{n+1}^3-a_n^3}{2}=-\frac{1}{2}a_n^3+\frac{1}{2}a_{n+1}^3
f(n)
f(1)+f(2)+f(3)+\cdots +f(60)
=\left(-\frac{1}{2}a_1^3+\frac{1}{2}a_2^3\right)\quad +\left(-\frac{1}{2}a_2^3+\frac{1}{2}a_3^3\right)\quad +\left(-\frac{1}{2}a_3^3+\frac{1}{2}a_4^3\right)\quad +\cdots +\quad \left(-\frac{1}{2}a_{60}^3+\frac{1}{2}a_{61}^3\right)
=-\frac{1}{2}a_1^3+\frac{1}{2}a_{61}^3=-\frac{1}{2}\left(\sqrt{2-1}\right)^3+\left(\frac{1}{2}\sqrt{2\cdot 61 - 1}\right)^3
=\frac{1}{2}(\sqrt{121}^3-1)=\frac{1}{2}(1331-1)=665
p_1=1
p_2=1
p_{n+2}=p_{n+1}+p_n\left(n \ge 1\right)
\left{p_n\right}
p_n=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right\}
X_n\left(n=1,2,\cdots\right)
X_1=1
X_n
\alpha
\alpha
\alpha
\alpha
X_n
X_{n+1}
X_1=1
X_2=10
X_3=101
X_4=10110
X_5=10110101
X_n
a_n
X_n
b_n
b_1=0
b_2=0
b_3=1
b_4=1
b_5=3
X_n
I_n
O_n
a_n=I_n+O_n
\left{\begin{array}{l}I_n=I_{n-1}+O_{n-1}\\O_n=I_{n-1}\end{array}
O_{n-1}
I_n
\left{\begin{array}{l}I_n=I_{n-1}+I_{n-2}\\O_n=I_{n-1}\end{array}
I_n
I_1=1
I_2=1
I_n=p_n=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right\}
O_n=I_{n-1}=p_{n-1}=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\right\}
a_n
a_n=I_n+O_n=p_n+p_{n-1}=p_{n+1}
a_n=p_{n+1}=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right\}
b_n
p_n
b_n
p_n
X_n
X_n
X_{n-1}
X_{n-2}
X_{n-1}
b_n=\left\{\begin{array}{lcl}b_{n-1}+b_{n-2}+1&:&n=odd\\b_{n-1}+b_{n-2}&:&else\end{array}
(n\ge3)
c_n
c_n=\left\{\begin{array}{ccl}1&:&n=odd\\0&:&else\end{array}
b_n
b_n=b_{n-1}+b_{n-2}+c_n
(n\ge3)
b_n
P_n
b_n
p_{n-1}
X_n
b_n
b_n=\left\{\begin{array}{lcl}p_{n-1}-1&:&n=even\\p_{n-1}&:&else\end{array}
(n\ge2)
c_n
b_n=p_{n-1}-1+c_n
(n\ge2)
d_n=1-c_n
b_n=p_{n-1}-d_n
(n\ge2)
b_1=0
b_2=p_1-d_2=1-(1-c_n)=1-1+0=0
b_3=p_2-d_3=1-(1-c_n)=1-1+1=1
b_n=p_{n-1}-d_n
b_{n-1}=p_{n-2}-d_{n-1}
b_{n+1}
b_{n+1}=(b_n)+(b_{n-1})+c_{n+1}=(p_{n-1}-d_n)+(p_{n-2}-d_{n-1})+c_{n+1}
=(p_{n-1}+p_{n-2})+(-d_n-d_{n-1}+c_{n+1})=p_n+(-d_n-d_{n-1}+c_{n+1})
-d_n-d_{n-1}=-1
b_{n+1}=p_n+(-d_n-d_{n-1}+c_{n+1})=p_n+(-1+c_{n+1})
d_n=1-c_n
b_{n+1}=p_n+(-1+c_{n+1})=p_n+(-d_{n+1})=p_n-d_{n+1}
b_n=p_{n-1}-d_n
n
n+1
b_n=p_{n-1}-d_n=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\right\}-d_n
b_n=p_{n-1}-d_n
(n\ge2)
E=\sum_{k=1}^{n+m}p\(k\)k
E=1 \times \frac{1}{2}+2\times \frac{1}{2}=1.5
E=\sum_{k=1}^{m+1}k\frac{1}{m+1}=\frac{1}{m+1}\cdot \frac{1}{2}(m+1)(m+2)=\frac{1}{2}(m+2)
=\frac{1}{2}m+1
E=\sum_{k=2}^{m+2}p\(k\)k=\frac{1}{10}2+\frac{2}{10}3+\frac{3}{10}4+\frac{4}{10}5=\frac{2+6+12+20}{10}=4
E=\frac{1}{10}2+\frac{2}{10}3+\frac{3}{10}4+\frac{4}{10}5=\frac{1}{10}(1\cdot 2 + 2\cdot 3 + 3\cdot 4+4\cdot 5)=4
M=\frac{(m+1)(m+2)}{2}
k\cdot (k-1)
E=\frac{1}{M}\sum_{k=2}^{m+2}k\cdot(k-1)=\frac{2}{(m+1)(m+2)}\sum_{k=2}^{m+2}k\cdot(k-1)
E=\frac{1}{M}\sum_{k=2}^{m+2}k\cdot(k-1)\\=\frac{2}{(m+1)(m+2)}\sum_{k=2}^{m+2}k^2-k=\frac{2}{(m+1)(m+2)}(\sum_{k=2}^{m+2}k^2-\sum_{k=2}^{m+2}k)\\=\frac{2}{(m+1)(m+2)}((\sum_{k=1}^{m+2}k^2-\sum_{k=1}^{1}k^2)-(\sum_{k=1}^{m+2}k-\sum_{k=1}^{1}k))
\Sigma
E=\frac{2}{(m+1)(m+2)}((\frac{1}{6}(m+2)(m+3)(2m+5)-1)-(\frac{1}{2}(m+2)(m+3)-1))\\=\frac{2}{(m+1)(m+2)}(\frac{1}{6}(m+2)(m+3)(2m+5)-\frac{1}{2}(m+2)(m+3))
=\frac{1}{3(m+1)(m+2)}((m+2)(m+3)(2m+5)-3(m+2)(m+3))\\=\frac{(m+2)(m+3)(2m+2)}{3(m+1)(m+2)}=\frac{2(m+2)(m+3)(m+1)}{3(m+1)(m+2)}=\frac{2(m+3)}{3}\\E=\frac{2(m+3)}{3}=\frac{2m}{3}+2
E=\frac{2m}{3}+2=\frac{2\cdot 3}{3}+2=2+2=4
n=1
\frac{1}{2}m+1
n=2
\frac{2}{3}m+2
E=\frac{n}{n+1}m + n
E=\sum_{k=n}^{n+m}p\(k\)k
[n,n+m]
p(k)
p(k)
{}_{n+m} \mathrm{C}_{n}
{}_{a+b} \mathrm{C}_{a}
{}_{n+m+1} \mathrm{C}_{n+1}
{}_{n+m-1}\mathrm{C}_{n-1}
{}_{n+m}\mathrm{C}_{n}
p(k)
p(k)=\frac{{}_{k-1}\mathrm{C}_{n-1}}{{}_{n+m}\mathrm{C}_{n}}
E=\sum_{k=n}^{n+m}p\(k\)k\\=\sum_{k=n}^{n+m}\frac{{}_{k-1}\mathrm{C}_{n-1}}{{}_{n+m}\mathrm{C}_{n}}\cdot k
{}_a \mathrm{C}_b = \frac{a!}{(a-b)!b!}
E=\sum_{k=n}^{n+m}\frac{{}_{k-1}\mathrm{C}_{n-1}}{{}_{n+m}\mathrm{C}_{n}}\cdot k\\=\sum_{k=n}^{n+m}\frac{m!n!}{(n+m)!}\frac{(k-1)!}{(n-1)!(k-n)!}\cdot k\\=\frac{m!n!}{(n+m)!}\sum_{k=n}^{n+m}\frac{(k-1)!}{(n-1)!(k-n)!}\cdot k\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}\frac{(k-1)!}{(k-n)!}\cdot k
\Sigma
(k-1)!k=k!
\frac{k!}{(k-n)!}={}_k \mathrm{P}_n
E=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}\frac{(k-1)!}{(k-n)!}\cdot k\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}\frac{k!}{(k-n)!}\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}{}_k\mathrm{P}_n
\sum_{k=n}^{n+m}{}_k\mathrm{P}_n
\sum_{k=1}^{a}{}_k \mathrm{P}_b=\frac{1}{(b+1)}\overbrace{(a+1)a(a-1)\cdots(a+1-b)}^{b+1}
E=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\sum_{k=n}^{n+m}{}_k\mathrm{P}_n\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}(\sum_{k=1}^{n+m}{}_k\mathrm{P}_n-\sum_{k=1}^{n-1}{}_k\mathrm{P}_n)\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}(\frac{1}{(n+1)}(n+m+1)(n+m)\cdots(m+1)-\frac{1}{(n+1)}n(n-1)\cdots(n-n))\\=\frac{m!n!}{(n+m)!}\frac{1}{(n-1)!}\frac{1}{(n+1)}\frac{(n+m+1)!}{m!}\\=\frac{m!n!}{(n-1)!m!}\frac{(n+m+1)!}{(n+m)!}\frac{1}{(n+1)}\\=\frac{n(n+m+1)}{(n+1)}=\frac{nm}{(n+1)}+\frac{n(n+1)}{(n+1)}=\frac{n}{n+1}m+n
\frac{1}{(n+1)}n(n-1)\cdots(n-n)
(n-n)=0
E=\frac{n}{n+1}m + n
\sum_{k=1}^{a}{}_k \mathrm{P}_b=\frac{1}{(b+1)}\overbrace{(a+1)a(a-1)\cdots(a+1-b)}^{b+1}
E=\frac{10}{11}m + 10
p
n
(1+n)^p -n^p -1
p
(1+n)^p -n^p -1
= ({}_pC_01^p + {}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1} + {}_pC_pn^p) \quad - n^p \quad - 1
(1+n)^p -n^p -1
= (1 + {}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1} + n^p) \quad - n^p \quad - 1
= {}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1}
{}_pC_1n + \ldots + {}_pC_{p-1}n^{p-1}
=\sum^{p-1}_{k=1}\left({}_pC_k1^{p-k}n^{k}\right)
=\sum^{p-1}_{k=1}\left({}_pC_kn^{k}\right)
{}_pC_kn^{k}=pm
m
p
{}_pC_kn^{k}
{}_pC_kn^{k}
=\frac{p!}{(p-k)!}\frac{1}{k!}n^k
k \ge 1
p
k \ge 1
p
{}_pC_kn^{k}
=pm
m
(1+n)^p -n^p -1
\begin{array}{crlllllll}p=1:&1&+n&&&&&-n&-1\\p=2:&1&+2n&+n^2&&&&-n^2&-1\\p=3:&1&+3n&+3n^2&+n^3&&&-n^3&-1\\p=4:&1&+4n&+6n^2&+4n^3&+n^4&&-n^4&-1\\p=5:&1&+5n&+10n^2&+10n^3&+5n^4&+n^5&-n^5&-1\\\end{array}
p=4
=4
{}_pC_k
=\frac{p!}{(p-k)!}\frac{1}{k!}
k
0\le k \le
p
1\le k \le p-1
(1+n)^p -n^p -1
=pm
(1+n
p
f(x)=\frac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}
f(1)+f(2)+f(3)+\cdots +f(60
a_n=\sqrt{2n-1}
a_{n+1}\cdot a_n=\sqrt{2n+1}\sqrt{2n-1}=\sqrt{4n^2-1}
a_{n+1}^2+a_n^2=2n+1+2n-1=4n
a_{n+1}^2-a_n^2=2n+1-2n-1=2
f(n
(a_{n+1}-a_n)
f(n)=\frac{(a_{n+1}-a_n)(a_{n+1}^2+a_n^2+a_{n+1}\cdot a_n)}{(a_{n+1}-a_n)(a_{n+1}+a_n)}=\frac{a_{n+1}^3+a_{n+1}^2a_n+a_{n+1}a_n^2-a_{n+1}^2a_n-a_{n+1}a_n^2-a_n^3}{a_{n+1}^2-a_n^2}
=\frac{a_{n+1}^3-a_n^3}{2}=-\frac{1}{2}a_n^3+\frac{1}{2}a_{n+1}^3
f(n)
f(1)+f(2)+f(3)+\cdots +f(60)
=\left(-\frac{1}{2}a_1^3+\frac{1}{2}a_2^3\right)\quad +\left(-\frac{1}{2}a_2^3+\frac{1}{2}a_3^3\right)\quad +\left(-\frac{1}{2}a_3^3+\frac{1}{2}a_4^3\right)\quad +\cdots +\quad \left(-\frac{1}{2}a_{60}^3+\frac{1}{2}a_{61}^3\right)
=-\frac{1}{2}a_1^3+\frac{1}{2}a_{61}^3=-\frac{1}{2}\left(\sqrt{2-1}\right)^3+\left(\frac{1}{2}\sqrt{2\cdot 61 - 1}\right)^3
=\frac{1}{2}(\sqrt{121}^3-1)=\frac{1}{2}(1331-1)=665
p_1=1
p_2=1
p_{n+2}=p_{n+1}+p_n\left(n \ge 1\right)
\left{p_n\right}
p_n=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right\}
X_n\left(n=1,2,\cdots\right)
X_1=1
X_n
\alpha
\alpha
\alpha
\alpha
X_n
X_{n+1}
X_1=1
X_2=10
X_3=101
X_4=10110
X_5=10110101
X_n
a_n
X_n
b_n
b_1=0
b_2=0
b_3=1
b_4=1
b_5=3
X_n
I_n
O_n
a_n=I_n+O_n
\left{\begin{array}{l}I_n=I_{n-1}+O_{n-1}\\O_n=I_{n-1}\end{array}
O_{n-1}
I_n
\left{\begin{array}{l}I_n=I_{n-1}+I_{n-2}\\O_n=I_{n-1}\end{array}
I_n
I_1=1
I_2=1
I_n=p_n=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right\}
O_n=I_{n-1}=p_{n-1}=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\right\}
a_n
a_n=I_n+O_n=p_n+p_{n-1}=p_{n+1}
a_n=p_{n+1}=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right\}
b_n
p_n
b_n
p_n
X_n
X_n
X_{n-1}
X_{n-2}
X_{n-1}
b_n=\left\{\begin{array}{lcl}b_{n-1}+b_{n-2}+1&:&n=odd\\b_{n-1}+b_{n-2}&:&else\end{array}
(n\ge3)
c_n
c_n=\left\{\begin{array}{ccl}1&:&n=odd\\0&:&else\end{array}
b_n
b_n=b_{n-1}+b_{n-2}+c_n
(n\ge3)
b_n
P_n
b_n
p_{n-1}
X_n
b_n
b_n=\left\{\begin{array}{lcl}p_{n-1}-1&:&n=even\\p_{n-1}&:&else\end{array}
(n\ge2)
c_n
b_n=p_{n-1}-1+c_n
(n\ge2)
d_n=1-c_n
b_n=p_{n-1}-d_n
(n\ge2)
b_1=0
b_2=p_1-d_2=1-(1-c_n)=1-1+0=0
b_3=p_2-d_3=1-(1-c_n)=1-1+1=1
b_n=p_{n-1}-d_n
b_{n-1}=p_{n-2}-d_{n-1}
b_{n+1}
b_{n+1}=(b_n)+(b_{n-1})+c_{n+1}=(p_{n-1}-d_n)+(p_{n-2}-d_{n-1})+c_{n+1}
=(p_{n-1}+p_{n-2})+(-d_n-d_{n-1}+c_{n+1})=p_n+(-d_n-d_{n-1}+c_{n+1})
-d_n-d_{n-1}=-1
b_{n+1}=p_n+(-d_n-d_{n-1}+c_{n+1})=p_n+(-1+c_{n+1})
d_n=1-c_n
b_{n+1}=p_n+(-1+c_{n+1})=p_n+(-d_{n+1})=p_n-d_{n+1}
b_n=p_{n-1}-d_n
n
n+1
b_n=p_{n-1}-d_n=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\right\}-d_n
b_n=p_{n-1}-d_n
(n\ge2)

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Last-modified: 2012-09-04 (火) 14:08:00 (3422d)