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中山诗题
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2025/10/15 08:23
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中山诗题
求
1
2
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\begin{aligned} \cfrac{1}{2^{n+m}}\sum^{n}_{a=0}\sum^{m}_{b=0}\sum^{a+b}_{i=0}2i\dbinom{a}{i}\dbinom{b}{i} \end{aligned}
2
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m
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2
i
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(
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我们提取出
2
−
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n
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m
−
1
)
2^{-(n+m-1)}
2
−
(
n
+
m
−
1
)
，根据
oi-wiki 二项式推论第十条
则有
∑
a
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i
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=
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min
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i
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i
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=
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\begin{aligned} \sum^{n}_{a=0}\sum^{m}_{b=0}\sum^{a+b}_{i=0}i\dbinom{a}{i}\dbinom{b}{i} &=\sum^{n}_{a=0}\sum^{m}_{b=0}\sum^{\min(a,b)}_{i=1}i\dbinom{a}{i}\dbinom{b}{i}\\ &=\sum^{\min(n,m)}_{i=1}i \left( \sum^{n}_{a=0}\dbinom{a}{i}\right)\left(\sum^{m}_{b=0}\dbinom{b}{i}\right) \\ &=\sum^{n}_{i=1}i \dbinom{n+1}{i+1}\dbinom{m+1}{i+1} \end{aligned}
a
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=
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1
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n
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i
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=
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i
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a
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0
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n
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=
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n
i
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i
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(
i
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)
我们令
j
=
i
+
1
,
i
=
j
−
1
j=i+1,i=j-1
j
=
i
+
1
,
i
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j
−
1
，则有
∑
i
=
1
n
i
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n
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1
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(
m
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=
∑
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j
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(
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m
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=
∑
j
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1
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j
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(
n
+
1
j
)
(
m
+
1
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)
+
1
=
∑
j
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1
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+
1
j
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)
(
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−
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\begin{aligned} \sum^{n}_{i=1}i \dbinom{n+1}{i+1}\dbinom{m+1}{i+1} &=\sum^{n+1}_{j=2}(j-1) \dbinom{n+1}{j}\dbinom{m+1}{j}\\ &=\sum^{n+1}_{j=0}(j-1) \dbinom{n+1}{j}\dbinom{m+1}{j}+1\\ &=\sum^{n+1}_{j=1}j \dbinom{n+1}{j}\dbinom{m+1}{j}-\sum^{n+1}_{j=0}\dbinom{n+1}{j}\dbinom{m+1}{j}+1\\ \end{aligned}
i
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1
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i
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i
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=
j
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(
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=
j
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1
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(
j
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(
j
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+
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=
j
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1
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+
1
j
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j
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(
j
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1
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−
j
=
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n
+
1
(
j
n
+
1
)
(
j
m
+
1
)
+
1
根据
oi-wiki 范德蒙德卷积第四条
，和
oi-wiki 二项式推论第三条
，
∑
j
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n
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1
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n
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1
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)
(
m
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=
(
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2
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\begin{aligned} \sum^{n+1}_{j=0}\dbinom{n+1}{j}\dbinom{m+1}{j} &=\dbinom{n+m+2}{n+1}\\ &=\dbinom{n+m+1}{n+1}+\dbinom{n+m+1}{n} \end{aligned}
j
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0
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1
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j
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(
j
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=
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m
+
2
)
=
(
n
+
1
n
+
m
+
1
)
+
(
n
n
+
m
+
1
)
根据
oi-wiki 二项式推论第二条
，我们有
j
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=
(
n
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1
)
(
n
j
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)
\begin{aligned} j \dbinom{n+1}{j}=(n+1) \dbinom{n}{j-1} \end{aligned}
j
(
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=
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(
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令
k
=
j
−
1
,
j
=
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+
1
k=j-1,j=k+1
k
=
j
−
1
,
j
=
k
+
1
，和
oi-wiki 范德蒙德卷积第四条
∑
j
=
1
n
+
1
j
(
n
+
1
j
)
(
m
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1
j
)
=
(
n
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1
)
∑
j
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1
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(
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=
(
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∑
k
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n
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(
m
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=
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(
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\begin{aligned} \sum^{n+1}_{j=1}j \dbinom{n+1}{j}\dbinom{m+1}{j} &=(n+1)\sum^{n+1}_{j=1}\dbinom{n}{j-1}\dbinom{m+1}{j}\\ &=(n+1)\sum^{n}_{k=0}\dbinom{n}{k}\dbinom{m+1}{k+1}\\ &=(n+1)\dbinom{n+m+1}{n+1} \end{aligned}
j
=
1
∑
n
+
1
j
(
j
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+
1
)
(
j
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+
1
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=
(
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j
=
1
∑
n
+
1
(
j
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1
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(
j
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1
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=
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k
=
0
∑
n
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=
(
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则
A
n
s
=
(
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+
1
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(
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m
+
1
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+
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)
−
(
(
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+
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+
1
=
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−
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+
1
\begin{aligned} Ans&= (n+1)\dbinom{n+m+1}{n+1}-\left(\dbinom{n+m+1}{n+1}+\dbinom{n+m+1}{n}\right)+1\\ &=n\dbinom{n+m+1}{n}-\dbinom{n+m+1}{n}+1 \end{aligned}
A
n
s
=
(
n
+
1
)
(
n
+
1
n
+
m
+
1
)
−
(
(
n
+
1
n
+
m
+
1
)
+
(
n
n
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+
1
)
)
+
1
=
n
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n
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+
1
)
−
(
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+
1
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