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19:T12ba,普遍的なことだけを説明しても中々伝わりづらいと思うので、具体的に問題を1問出しながら説明させてください！

まず前提として、応用の問題が解けるようになるためには以下のことが必要になります。(結論です)

・基本的な解法がすぐに出てくるようにする
・問題を見た時、前の問題との関連性から考えていく
・誘導に乗っていくのに慣れるのにはとにかく演習量が必要

1つ目は恐らく大丈夫だと思います。また、3つ目もこれから2次試験向けの演習を重ねるうちに「あの時の誘導に似てるなー」というような感覚で段々できるようになってくるものです。つまりは慣れです。自分自身もこれを強く感じています。最初は中々誘導に乗れず辛いかもしれませんが、まずは量をこなしましょう。

おそらく問題は2つ目です。
これは分かりやすく言うと、「こうやってやっていって…あ、(1)(2)ここで使う？」という考え方ではなく、「(1)や(2)の問題の考え方を上手く使えないかな〜」「今までやったことのある基本問題の考え方が何か使えないかな〜、あ、文章のこの部分前にやったあの問題文と似てるな〜」と言ったような、初めから誘導や基本問題などのヒントの方から答えを探っていくように考えていくことです(長くてごめんなさい)。

実際に問題を見て考えていきましょう！以下は2015年の九大の問題です。

以下の問いに答えよ。
(1)nが正の偶数のとき、2^n-1は3の倍数であることを示せ。
(2)pを素数とし、kを0以上の整数とする。2^(p-1)-1=p^kを満たすp,kの組を全て求めよ。
(※^の後は指数を表します。2^n-1は2のn乗-1、2^(p-1)-1は2のp-1乗-1です)

(1)は割愛しますが、n=2l(lは自然数)とかと置いて二項定理で分解して3で括ったり、帰納法を使えばいいと思います。とにかく2^n-1が3の倍数だと分かればいいです。

問題は(2)ですね。先程言った通り、誘導を上手く使えないかという点からとにかく問題を見ましょう！
まず見るべき点は式の形が左辺と似ている所です。誘導が使えそうですよね。

誘導を上手く使うコツですが、「誘導の部分と問題文の該当部分の違いを上手く見分けること」です。今回であればnがp-1に変わっています。また、(1)でnは"正の偶数"でしたが、p-1は"素数-1"ですよね。

ここの違いは何かあるでしょうか？？

まず整数問題で素数が出たら、「2とそれ以外」という見方をするのは演習量をこなせば分かってきます。素数の中でも2だけ偶数で稀有、と認識できていればOKです。(ここは基本問題的な解法暗記の部分)

素数-1は、素数が2のときだけ奇数、素数が2以外のときは偶数になりますよね！

ですので、2か2じゃない素数かで分けます。2じゃない素数のときは(1)の条件と一致するので使えそうですよね。まずは使いましょう！

○pが2以外の素数のとき
(1)より左辺は3の倍数です。ということは右辺も3の倍数になります。p^k、つまり素数の累乗が3の倍数ということはpは3以外ありえないですよね。ここは素数ならではです。
ですのでp=3から左辺に代入するとk=1と決まります。

○pが2のとき
代入していくとk=0になりますね。

以上から(p,k)=(3,1),(2,0)となりました！

このように、「基本問題の解法はすぐに出ておくようにする」「誘導から常に考えていく(誘導と問題文の違いを認識し、見分けていく)」ことの重要性がわかったと思います。また、基本問題というのは、教科書や青チャートにある典型問題もそうですが、素数は2とそれ以外に分ける、といったような"応用問題でよく出てくるテクニック"もそうです！これは演習量を詰まないと中々インプットされないので、「演習量が大切」なのも再認識できるでしょう。

また、1問に時間をかけて思考していくこともとても大切です！最終的にその標準問題の解き方を覚えられると役には立ちますが、思考力というのは思考する時間を取らないと中々伸びません。1問に10分は考える時間を取りましょう！

めちゃくちゃ長くなって申し訳ないですが、参考になれば幸いです！！

1b:T1307,　確かに解法暗記は大切です。しかし、それを単純暗記で終わらせてしまっては危険です。京大の整数問題を例に見ていきましょう。

「n^3ー7n＋9が素数となるような整数nを全て求めよ。」（2018）
　この問題は、整数kを用いて、nを3k、3k＋1、3kー1とに場合分けして考えればすぐ解けます。しかし、この解法を単純に暗記しても、どこからこの解法を導く着想を得たのかが分からなければ、同じ解法を使う問題に対峙してもそれを見抜くことは困難です。この問題では、n＝1を仮に入れてみると、値は3で素数です。次に、n＝2を入れてみた場合、こちらも値は3で素数です。n＝3の場合は15で素数ではない、n＝4の場合は45で素数ではない、n＝5の場合は99で素数ではない……。ここで何か気づくでしょう。すなわち、実験して得られた値は全て3の倍数になっていることに気づくはずです。となれば、与式の取りうる値は全部3の倍数なんじゃないか？という疑いが生じるでしょう。この仮説を確かめるために、まずはすべてのnに対し与式の値は必ず3の倍数になるということを証明すればよいことになり、そのためにnを3で割った余りに注目して場合分けをするという解法に辿り着くわけです（したがって、modを使えばもっと楽な計算で証明できます）。(i)n＝3kの場合は言うまでもないとして、(ii)n＝3k＋1の場合、与式は27k^3＋27k^2ー12k＋3で、(iii)n＝3kー1の場合、27k^3ー27k^2ー12k＋15で、いずれも3の倍数になります。素数の中で3の倍数は3だけなので、結局この問題は、（与式）＝3という方程式を整数nについて解けば良いということになります。
　
　こんな感じで解法を深く見つめていくと、解ける問題も増えていきます。例えば、この問題。
「pが素数ならばp^4＋14は素数でないことを示せ。」（2021文系）
　p＝2のとき値は30、p＝3のとき値は95、p＝5のとき値は639、p＝7のとき値は2415、p＝11のとき値は14655……。p＝3のとき以外は、いずれも3の倍数です。よって、(i)p＝3のときと、(ii)p ≠ 3の時で場合分けをして、(ii)p ≠ 3のときでは、さらに(a)p ≡ 1（mod3）のときと、(b)p ≡ 2（mod3）のときとで場合分けして、p^4＋14が素数pに対し常に3の倍数となることを証明し、そのとき取りうる値は3のみであるが、p^4＋14はp＝2で最小値30であるから、3を取ることはない。したがって、p^4＋14は素数ではない、という解決ができるわけです。
　
　また、この問題も。
「素数p, qを用いて、p^q＋q^pと表される素数をすべて求めよ。」（2016理系）
　pとqの対称性からp≦qとしても一般性は失われないので、この大小関係のもと進めていきます。まず、2数の偶奇が一致するとき、その和は必ず偶数になりますが、pとqはいずれも素数なので、与式の取りうる値は最小でも8（p＝2, q＝2）であり、値が2となることはありません。このことから、与式の値は奇数であり、そのためにはp＝2でなければなりません（片方は偶数でなければならず、p^qが偶数となるのはp＝2の場合だけ）。すると、p＝2と固定して、qに3、5、7、11……と入れてみればいいわけです。q＝3のとき値は17で素数、q＝5のとき値は57で素数ではない、q＝7のとき値は177で素数ではない、q＝11のとき値は2169で素数ではない……。q＝3のときを除いて、すべて3の倍数ですね。しかし、この問題では、安易にqを3で割った余りで場合分けしてもうまくいきません。場合分けにさらなる工夫が必要になりますが、そこは自力でやってみましょう。

　上の問題は、いずれも同じところから解法の着想を得ていることがわかったと思います。と同時に、個別の問題にだけ通用するような覚え方をしても、似た問題ですら手が止まってしまうということも。やはり何事も、勉強というからには自分の頭で考えなければなりません。ただ単に、与えられた結果の知識や表現を覚えるだけではダメですね。その点、受験勉強は大変なものですが、そういったことも志望校という目標に向かって一途に続けられる人こそ、本番で勝っていく人たちなのでしょう。私も偉そうなことは言えませんがね。1c:Te31,こんにちは！
たしかに三元一次は煩雑になってミスりがちですね笑
自分もベクトルの大きさの計算なんかはかなり苦手でした。

以下、計算ミスを防ぐために（特に共通テストで）気をつけるポイントをお伝えします！



①ベクトルの成分は縦に書く
もしかしたら既にやっているかもしれませんが、ベクトルの成分は縦に並べて書きましょう。現行の教科書などは成分が横『（２，4，3）のような形』で書かれていることが多いですが、これだとミスりやすいです。
　　2
｛　4　｝
　　1
のように縦で成分表示すると文字が入って式が複雑になっても見やすいので、成分同士の方程式や内積の計算をするときのミスがかなり減ります。

(OP→)＝x(a→) ＋y(b→) + z(c→)
のような場合も、
　　　　　　ａ
(OP→)＝｛　ｂ　｝
　　　　　　ｃ
のように表しちゃうと計算でミスりづらいです！


②大きな余白や白紙のページを利用する
共通テスト本番ではめちゃくちゃ煩雑なベクトルの計算が出ることは正直あまりないです。しかし、東進などの予備校が手掛けている模試や問題集の中には、計算ゲーのような悪問も含まれているのが現状です。ですので正直に言えば、そういった模試などの悪問でケアレスミスをしてしまっても一喜一憂することは無いと思います。
しかし、工夫をするとすればやはり余白の使い方でしょう。「あ、この計算重いわ」と感じたら、無理して小さい余白や暗算に頼らず、どっしりと構えて大きな余白を探しましょう。その分タイムロスに感じるかもしれませんが、いくらわさんのように京大を目指すレベルであれば、タイムロスよりも安易な判断による失点の方が痛いことは明確だと思います。心に余裕を持って頑張ってください！


③後回しにする
共通テストの数学は、ひらめきゲー／誘導ゲーな要素があります。自分のやり方でやったら死ぬほど難しい式がでてきたけど，誘導にうまく乗っかって解き直したらめちゃくちゃ簡単だった、なんてケースがかなり多いです。また、わからないからとりあえず飛ばして最後に戻ってきたら、頭がクリアになって簡単に解けたというケースも多いです。
問題が変に難しいなと感じた時は、割り切ってスキップして、最後に戻ってくるようにしましょう。仮に計算ミスをしていたとしても、後で見直すと間違いに気づきやすいです。共通テストはとにかく時間と勝負なので、沼りはじめたら終わります。とりあえずスキップしてみることは案外大切な心構えですよ！


①〜③までご紹介しましたが、特に大事なのは③です。
これは共通テストの数学では本当に大切な考え方です！一緒に受験勉強していた東大生の友人たちでさえ、計算が煩雑になったり沼ったりすることがありましたし、そういう時はとりあえず飛ばして最後に戻ってくるのがいいと話していました。

ぜひ参考にしてください！
また、これから過去問などで形式に慣れていけば、だんだん計算ミスは減ってくると思いますよ〜！頑張ってください！2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=7wzMPK7kmp1o62NQGS9N","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"符号の統一の仕方"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",2,") コメント(",0,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"8/7 0:10"}]]}],["$","div",null,{"className":"coach-mark 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mb-1","children":"質問者様は高2ということなので、数Ⅱまでの範囲で回答させていただきます。\n\n\n【三角関数を変形する目的】\n\nまず、三角関数を変形するのは必ず目的があります。\n①三角関数を含んだ方程式・不等式を解くため\n②三角関数を含んだ関数の最大値・最小値を求めるため\nなどがよくある目的ですね。\n\n《①について》\n方程式や不等式ははじめに因数分解で攻めます。\n(因数)(因数)=0\nといった形になれば、あとは簡単ですね。\n因数分解しない場合は②の考え方をそのまま借りましょう\n\n《②について》\nsinのみ、cosのみ、tanのみ、の式に帰着させます。そしたら見たことある関数(一次関数、二次関数など)になります。\nそのための手段として\n＊三角関数の相互関係\n＊加法定理を用いた公式\nなどが存在します。\n\n\n－－－－－－－－－\n\n【質問主様の弱点と思われるところ】\n\n数Ⅱの三角関数に入ってからうまくいかなくなった高校生は加法定理を用いた公式につまづいている人が多いです。\n公式自体覚えていても、問題でうまく活用出来ないことがよくあります。\n\n先程の項目で書きました、変形のそもそもの目的を意識して演習してみてください。\n使い分けパターンは青チャートなどのテキストに詳しく記載されています。これを身につけることが大切です。\n\nパターンを繰り返しの演習で身につける際に、\n「因数分解を目指す！」\n「sinのみ、cosのみ、tanのみの式を目指す！」\nという意識を持って取り組むことで、何故その式変形を使うのかが体感出来ます。\n\n\n－－－－－－－－－\n\n【最後に】\n\n問題のゴールから逆算して考えることが数学においては大切です。\n初めから逆算して考えることなんて出来ないから、パターンを演習によって身につけるわけですが、ゴールを意識してパターンを身につけなければ、何のためのパターンなのかがわかりません。\n必ず、式変形の目的を意識した演習を心掛けてください。"}],["$","div",null,{"className":"flex 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"隣接3項間漸化式"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんにちは、名古屋大学医学部医学科のメイメイといいます。\n(an-an-1)=bnとするとb1は求められないですね。\n\n(an+1)-(an)=2［(an)-(an-1)］\nが出てきているはずですが、\n\nn-1の項があり基本的にn≧2で考えています。\nこれをn≧1に直してみると\n(an+2)-(an+1)=2［(an+1)-(an)］\nとなります。\n単純にnの部分を1ずつずらしただけです。\n\nこの状態で(an+1)-(an)=bn\nと置いてみましょう。\n\nb1が求められるはずです。(ちなみにb2は必要ないです。)\n\nつまり(bn+1)=2(bn)、b1=(a2)-(a1)=8の等比数列に帰着しますね。\n\nこれを解くと、bn=8・2^n-1=2^n+2となります。(2^n-1は2のn-1乗という意味です。)\n\nすなわち、(an+1)-(an)=2^n+2\n\n両辺を2^n+1で割ると\n\n<(an+1)/2^n+1>-(1/2)<(an)/2^n>=2\n\nとなります。\n\n(an)/2^nをcnとすると、(cn+1)=(1/2)(cn)+2\n\nこれを変形して、(cn+1)-4=(1/2)<(cn)-4>\n\nつまり(cn)-4=(-7/2)・(1/2)^n-1=(-7)・(1/2)^n\n\nよってcn=4-7・(1/2)^n\n\nこの両辺に2^nをかけてan=4・2^n-7 (n≧1)\n\nとなります。\n分かりにくくてすいません！\n\n"}],["$","div",null,{"className":"flex 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"符号変換ミスが減らせない"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"僕の習っていた予備校講師の言ってた言葉が、\"計算をするから計算ミスをする\"というものです。\n数学は解法の選択によって計算量が大きく変わりますので、この言葉を念頭に解こうと思えば、力がつくんじゃないでしょうか。\n\nさらに、開き直ってしまえば、だれでも計算ミスというのはあるものなんです。大学入試において、単純な計算問題というのは出ません。まあ、積分しろ！みたいなのはありますけど、、、\n\n例えば、図形問題や写像の通過領域とかの問題でしたら、僕なら最初にお絵描きして答えを予想しちゃいます。\nあとはそれを論証するために計算したりするので、ミスにはすぐ気づけますね。\n感覚としては確率の問題で1を超えちゃってるから違うなってすぐ気づけるイメージです。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 3.31-2 6-2m0-12C9.79 4 8 5.79 8 8s1.79 4 4 4 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mb-1","children":"三味線さん、はじめまして。\n\nお気持ちはすごく分かります。\nたしかに解答の細かいところに疑問を持ったり、その都度公式を導出していると参考書の進むペースは遅くなってしまいますが、その分、質は高くなると思うので全然良いことだと思いますし、むしろそうするべきだと思います。\n\nよく言われる「数学は理解」という言葉は、なぜその公式を使ったのか、なぜその解法で解くのか、なぜその変換を行うのか、もっと細かいことで言うと、なぜその順に解答を記述するのかといったことを理解することです。\n\n「数学は暗記」という言葉もたまに聞きますが、これは単純に英単語みたいに暗記すると言うことではなくて、どうしてこの解法を使うのかを理解した上でどうゆう問題が出たらどの解法を使うのかを暗記すると言うことです。\n仮に理解の過程を飛ばして暗記だけすると、少し問題の形が変わっただけで解法が思い浮かばないということになってしまいます。\n\nそして理解を深めるためには、三味線さんのように細かいところにも疑問を持って問題を解くのが一番の近道です。公式は導出ができる方が理解度ははるかに上がりますし、たまにある公式の導出に基づいた問題なんかも出題されることもあります。\nまた質問文中のことで触れると、なぜ置換積分はこうゆう形でするのか、一次独立とは何か、解答に使われている言葉の意図、こういったことに疑問をもって考えるのはとても良いことだと思います。確認しても忘れてしまうのは人間なので仕方ないことで、確認してその時に理解したことをノートなんかに纏めておきましょう。次に同じような疑問が出た時にノートを見返すことで少しずつ定着して力になっていくはずです。\n私の場合だと2.3回では定着せず、5回とか10回その都度見返すことで定着し始めた感じだったので、忘れているから力になっていないと焦らずに、自分のペースで頑張ってください！\n\n応援しています☺️\n"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"やり直しの仕方"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"数学の問題をやり直す上で、解答や式変形を一字一句覚えるなんていうことがな必要ないことは言うまでもないことだとおもいます。\nなぜなら、数値、条件が全く同じ問題なんて人生でそう出会わないからです。\n\nでは、どうするのか？ということですが、僕が意識していた点はその問題の核となる部分を抽出し抽象化、一般化することです。\n\n要は1から10を得てほしいと言えばいいのでしょうか？\n\n\n具体的に説明すると、立体図形の問題で、ベクトルで解こうとしたけど、なかなか上手くいかなかった。\n\n\n解答にはベクトルによる解法が書かれておりその解法がなかなかテクニカルで簡潔である。\nしかし別解に座標を置いて計算でごり押しする解き方も書いてある。こちらの方法はなかなか、計算量が多そうだ。\n\n\nこういうことがあったとします。\n\nこういう時に、じゃあテクニカルな式変形を覚えようとしていてはなかなか数学力はつきません。\n\nこの問題の復習はいくつかやり方が考えられますが、この問題の核を抽出し一般化とは、以下のようなことです。\n\n1.確かにベクトルのやり方もいい。なので、頭に留めておこう。\n\n2.座標を置くやり方は計算量が多い一方、やっていることは素直である。なので、本当に思いつかなかったら、最終的に座標を置けばいいのではないか？\n\n3.角度といった条件は出来るだけベクトルで扱うのが良さそうだ。\n\n4.交線などは、座標を置き平面の方程式を立てて求めていくのが良さそうだ。\n\n\nなどなど得られることはたくさんあるはずです。\n\n\nこれはあくまで一例ですが、１つの問題から学べることは案外多いものです。\n無作為に問題数をこなすのではなく密度の濃い演習をこなすことをお勧めします！\n\nあくまで僕個人の意見ですので、何か参考になれば幸いです。\n"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"数学の解法を思い付くためには"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"　①問題文で与えられている情報と、②最終的に問われている内容とを区別して、①→②にどう持っていくかを考えることが、数学の問題を解く上での第一歩だと思います。しかし、無駄な思考要素で時間を必要以上に割いてしまうのも勿体無いので、実際に考えるときは、思考の道のりを最短化するために②→①の方向に逆算して考えましょう。②を求めるためにはaとbとcの方法があって、aの方法で解くとしたらa-1とa-2とa-3が分かればよくて、a-1が分かるために①の情報をどれか使えないか、a-2が分かるために①から必要な情報を何か導けないか、a-3が分かるためには①のどれをどう使えば良いか…という感じです。これが入試問題をはじめとする応用問題への取り組み方です。\n　そして、一般に基礎問題は、「②を求めるためにはどういった方法があるか」、「aの解法には何がわかっている必要があるか」、「①の情報からどうa-1を導くか」といった、細分化された要素のレベルに留まるものです。応用問題は基礎問題の組み合わせだとよく聞きますが、それはこういう意味です。なので、解法が思いつかない時は、「困難は分割せよ」の精神で、その問題を解くために必要となる過程を細分化してみると良いと思います。普段の勉強では、①と②を書き出して、②→a→a-1,a-2,a-3→①といった解答の設計図を書くように心がけると、必要な思考過程を分析する癖がついて、模試などでも役に立つかもしれませんね。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 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