第1回変なΣ 「連続3回」

$\sum_{1 \leq i < j < k \leq n} = (\binom{n}{3})$
$\sum_{1 \leq i \leq j \leq k \leq n} = ((\binom{n}{3}))$

テレスコで

$\begin{aligned} \sum_{1 \leq i < j < k \leq n} & = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} (\binom{i - 1}{0}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} [(\binom{i}{1}) - (\binom{i - 1}{1})] \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} (\binom{j - 1}{1}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} [(\binom{j}{2}) - (\binom{j - 1}{2})] \\ = \sum_{k = 3}^{n} (\binom{k - 1}{2}) \\ = \sum_{k = 3}^{n} [(\binom{k}{3}) - (\binom{k - 1}{3})] \\ = (\binom{n}{3}) . \end{aligned}$

$\begin{aligned} \sum_{1 \leq i \leq j \leq k \leq n} & = \sum_{k = 1}^{n} \sum_{j = 1}^{k} \sum_{i = 1}^{j} \\ = \sum_{k = 1}^{n} \sum_{j = 1}^{k} \sum_{i = 1}^{j} ((\binom{i}{0})) \\ = \sum_{k = 1}^{n} \sum_{j = 1}^{k} \sum_{i = 1}^{j} [((\binom{i}{1})) - ((\binom{i - 1}{1}))] \\ = \sum_{k = 1}^{n} \sum_{j = 1}^{k} ((\binom{j}{1})) \\ = \sum_{k = 1}^{n} \sum_{j = 1}^{k} [((\binom{j}{2})) - ((\binom{j - 1}{2}))] \\ = \sum_{k = 1}^{n} ((\binom{k}{2})) \\ = \sum_{k = 1}^{n} [((\binom{k}{3})) - ((\binom{k - 1}{3}))] \\ = ((\binom{n}{3})) . \end{aligned}$

理屈で

$\sum_{1 \leq i < j < k \leq n} = | {i, j, k | 1 \leq i < j < k \leq n} | = (\binom{n}{3}) .$

$\sum_{1 \leq i \leq j \leq k \leq n} = | {i, j, k | 1 \leq i \leq j \leq k \leq n} | = ((\binom{n}{3})) .$

$\sum_{1 \leq i < j < k \leq n} i = (\binom{n + 1}{4})$
$\sum_{1 \leq i < j < k \leq n} j = 2 (\binom{n + 1}{4})$
$\sum_{1 \leq i < j < k \leq n} k = 3 (\binom{n + 1}{4})$

$\begin{aligned} \sum_{1 \leq i < j < k \leq n} i & = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} i \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} i (\binom{i - 1}{0}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} 1 (\binom{i}{1}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} [(\binom{i + 1}{2}) - (\binom{i}{2})] \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} (\binom{j}{2}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} [(\binom{j + 1}{3}) - (\binom{j}{3})] \\ = \sum_{k = 3}^{n} (\binom{k}{3}) \\ = \sum_{k = 3}^{n} [(\binom{k + 1}{4}) - (\binom{k}{4})] \\ = (\binom{n + 1}{4}) . \end{aligned}$

$\begin{aligned} \sum_{1 \leq i < j < k \leq n} j & = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} j \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} j (\binom{i - 1}{0}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} j [(\binom{i}{1}) - (\binom{i - 1}{1})] \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} j (\binom{j - 1}{1}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} 2 (\binom{j}{2}) \\ = 2 \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} [(\binom{j + 1}{3}) - (\binom{j}{3})] \\ = 2 \sum_{k = 3}^{n} (\binom{k}{3}) \\ = 2 \sum_{k = 3}^{n} [(\binom{k + 1}{4}) - (\binom{k}{4})] \\ = 2 (\binom{n + 1}{4}) . \end{aligned}$

$\begin{aligned} \sum_{1 \leq i < j < k \leq n} k & = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} k \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} k (\binom{i - 1}{0}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} \sum_{i = 1}^{j - 1} k [(\binom{i}{1}) - (\binom{i - 1}{1})] \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} k (\binom{j - 1}{1}) \\ = \sum_{k = 3}^{n} \sum_{j = 2}^{k - 1} k [(\binom{j}{2}) - (\binom{j - 1}{2})] \\ = \sum_{k = 3}^{n} k (\binom{k - 1}{2}) \\ = \sum_{k = 3}^{n} 3 (\binom{k}{3}) \\ = 3 \sum_{k = 3}^{n} [(\binom{k + 1}{4}) - (\binom{k}{4})] \\ = 3 (\binom{n + 1}{4}) . \end{aligned}$