root-factorial-sum

検索用コード
{\underline{\textcolor{black}{\ruizyoukon2}}})+(\textcolor{red}{\teisei{\textcolor{black}{\ruizyoukon4}}}-\textcolor{red}{\teisei{\textcolor{black}{\ruizyoukon3}}})+(\textcolor{red}{\teisei{\textcolor{black}{\ruizyoukon5}}}-\textcolor{red}{\teisei{\textcolor{black}{\ruizyoukon4}}})+\cdots+(\textcolor{green}{\underline{\textcolor{black}{\ruizyoukon{64}}}}-\textcolor{red}{\teisei{\textcolor{black}{\ruizyoukon{63}}}})$ \\[.2zh] \phantom{ (1)}\ $\phantom{\therefore\ \bm{S_{62}}}=-\textcolor{green}{\underline{\textcolor{black}{\ruizyoukon2}}}+\textcolor{green}{\underline{\textcolor{black}{\ruizyoukon{64}}}}=\bm{-\ruizyoukon2+8}$ \\\\ \centerline{{\normalsize $\left[\textcolor{brown}{\begin{array}{l} \bm{有理化}すると階差の形になる. \\ k=1からk=62まで和の形で書き出すと,\ 2つの項だけが残る. \\ どの項が残ってどの項が消えるのかを慎重に判断する必要がある. \end{array}}\right]$}} \\\\\\\\  (2)\ \ $\textcolor{cyan}{k\cdot k\kaizyou}=\{(k+1)-1\}\cdot k\kaizyou=\textcolor{red}{(k+1)\kaizyou-k\kaizyou}$ \\[1zh] \phantom{ (2)}\ $\therefore \bm{S_n}=(\textcolor{red}{\teisei{\textcolor{black}{2\kaizyou}}}-\textcolor{green}{\underline{\textcolor{black}{1\kaizyou}}})+(\textcolor{red}{\teisei{\textcolor{black}{3\kaizyou}}}-\textcolor{red}{\teisei{\textcolor{black}{2\kaizyou}}})+(\textcolor{red}{\teisei{\textcolor{black}{4\kaizyou}}}-\textcolor{re \phantom{ (2)}\ $\phantom{\therefore \bm{S_n}}=\bm{\textcolor{green}{\underline{\textcolor{black}{1\vphantom{\bunsuu{1}{()\kaizyou}}}}}-\textcolor{green}{\underline{\textcolor{black}{\bunsuu{1}{(n+1)\kaizyou}}}}}$ \\\\\\ \centerline{{\normalsize $\left[\textcolor{brown}{\begin{array}{l} (2)と(3)は階差への変形を覚えているかどうかがものをいう. \\ ここで経験しておいてほしい.