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※高校数学Bの「数列」について,このサイトには次の教材があります.
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規則性を見つける
一般項に慣れる
等差数列
等比数列,循環数列
和の記号Σ
同 (2)
同 (3)
Σ記号の変形
等比数列のΣ
いろいろな数列のΣ
階差数列
Snan関係式
部分分数分解
等差×等比形の数列の和-現在地
群数列
自然数の累乗の和
センター試験.数B.数列

== 等差と等比の積になっている数列 ==

【要点】
・中間項が消えれば和が求まります。
・中間項が等比数列などになっても和が求まります。

○等比数列の和
等比数列の和は,次のようにして求めました。

(1) 求める和をSとおく。
(2) 公比rをかけてrSを作る。
(3) S-rSを求めると,中間項が消えて,Sが求まる。


S(1−r)=a-arn となるから

○等差数列と等比数列の積になっている数列の和
「等差数列と等比数列の積になっている数列」とは,例えば次のようなものをいいます.
【例1】
初項1,交差1の等差数列an=nと初項1,公比2の等比数列bn=2n−1の積
anbn=n×2n−1
初めの幾つかを書くと
1, 2×21, 3×22,..., r×2r−1,..., n×2n−1
【例2】
初項3,交差2の等差数列an=2n+1と初項3,公比3の等比数列bn=3nの積
anbn=(2n+1)3n
初めの幾つかを書くと
3×31, 5×32, 7×33,..., (2r+1)3r,..., (2n+1)3n

 等比数列の場合,S-rS により「中間項が消えるので」和が求まります.
 等差×等比型の場合,結果を覚える形にすることはできませんが,上で示した方法を真似すると 「中間項が消えなくても」「中間項の和が求まるので」全体の和が求まります。
この場合は,「中間項が等比数列」となり,和が求まることになります。
【要点】
S−rSを作る
Sが等比数列の和ならば
 「中間項が消えて」和が求まる.
Sが等差数列×等比数列の和ならば
 「中間項が等比数列になって」和が求まる.

【例1】 次の和を求めなさい。ただし,r≠1とする。
1+2r+3r2+4r3+5r4+・・・+nrn−1
答案
S=1+2r+3r2+4r3+5r4+・・・+nrn−1とおく
−)rS=r+2r2+3r3+4r4+・・・+(n-1)rn−1+nrn
S−rS=(r0+r1+r2+r3+r4+・・・+rn−1)−nrn

()内は等比数列の和になるので公式で和が求まります。
()内は初項1,公比等比r,項数nr1からrn−1まででn−1項,これに1=r0の項を付け加えるとn項)の等差数列の和数列の和だから
1(rn1)r1
結局,求めるものは
(1r)S=1(rn1)r1nrn=1rn1rnrn
S=1rn(1r)2nrn1r=rn(nrn1)(1r)2+1(1r)2 …(答)

(ビックリ答案から学ぼう)
次の数列は等比数列ではありません: S=1+2r+3r2+4r3+5r4+・・・+nrn−1
今までに出会った間違い答案で,「どうしても合わない」という人は,この数列の公比を
anan1=nrn1(n1)rn2=rnn1
と考えていた例がありました。公比は”公”なのでnによらない共通な定数でなければなりませんが,この数列には公比はありません!

どうしても納得しない方は,
1,3,5,7,… ,(2n−3),(2n−1)
という数列[公差2の等差数列]公比が
r=2n12n3
とは言えない
ことを思い出しましょう.

(参考)
wxMaxima(←インストール方法など)を使って,検算する方法
メニューから:微積分→数列の総和→
[関数]k*r^(k-1)
[変数]k
[下端]1
[上端]n
[数式で出力]にチェックを入れる
これにより,次の形で出力されます.
rn(nrn1)(1r)2+1(1r)2

【例2】 次の和を求めなさい。ただし,x≠1とする。 k=1n(k+1)xk
答案
S=2x+3x2+4x3+5x4+・・・+(n+1)xnとおく
−)xS=2x2+3x3+4x4+・・・+nxn+(n+1)xn+1
S−xS=2x+(x2+x3+x4+・・・+xn)−(n+1)xn+1

()内は等比数列の和になるので公式で和が求まります。
()内は初項x2,公比等比x,項数n−1の等差数列の和数列の和だから
x2(xn11)x1=x2(1xn1)1x

結局,求めるものは
(1x)S=2x+x2(1xn1)1x(n+1)xn+1
S=x2(1xn1)(1x)2+2x(n+1)xn+11x
=xn+1{(n+1)xn2}+2xx2(1x)2 …(答)

(参考)
wxMaximaを使って,検算する方法
メニューから:微積分→数列の総和→
[関数](k+1)*x^k
[変数]k
[下端]1
[上端]n
[数式で出力]にチェックを入れる
これにより,次の形で出力されます.
xn+1(nx+xn2)(x1)2(x2)x(x1)2

■問題
1 次の和を求めなさい。ただし,x≠1とする。
1+3x+5x2+7x3+・・・+(2n-1)xn-1
(答案)
(正しいものを選びなさい。)
 
[ア] (2n-1)xn-1, (2n-3)xn-12nxn-1, (2n+1)xn-1
[イ] 1x2x
[ウ] x2x2x2
[エ] n-1nn+1
 

■問題
2 次の和を求めなさい。
(答案)
(正しいものを選びなさい。)
 
[ア] n3n-13n(n-1)3nn3n
[イ] 1332
[ウ] x332
[エ] n-1nn+1


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