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== センター試験.数B.数列(2013~) ==
【2013年度センター試験.数学Ⅱ・B】第3問(選択問題)
(1) 数列{pn}は次を満たすとする。
p1=3,pn+1=13pn+1(n=1,2,3,)・・・①
数列{pn}の一般項と,初項から第n項までの和を求めよう。まず,①から
pn+1
=13(pn
)(n=1, 2, 3, ···)

となるので,数列{pn}の一般項は
pn=
1
n−2
+

である。したがって,自然数nに対して
k=1npk=
(1−
1
n
)+
n

である。

解説を読む

解説を読む

(2) 正の数からなる数列{an}は,初項から第3項がa1=3, a2=3, a3=3であり,すべての自然数nに対して
an+3=an+an+1an+2 ・・・②
を満たすとする。また,数列{bn}, {cn}を,自然数nに対して,bn=a2n−1, cn=a2nで定める。数列{bn}, {cn}の一般項を求めよう。まず,②から
a4=a1+a2a3=,a5=3,a6=
,a7=3

である。したがって,b1=b2=b3=b4=3となるので
bn=3 (n=1, 2, 3, ・・・)・・・③
と推定できる。
 ③を示すためには,b1=3から,すべての自然数nに対して
bn+1=bn ・・・④
であることを示せばよい。このことを「まず,n=1のとき④が成り立つことを示し,次に,n=kのとき④が成り立つと仮定すると,n=k+1のときも④が成り立つことを示す方法」を用いて証明しよう。この方法をという。に当てはまるものを,次の⓪~③のうちから一つ選べ。
⓪ 組立除法 ① 弧度法 ② 数学的帰納法 ③ 背理法

解説を読む

[Ⅰ] n=1のとき,b1=3, b2=3であることから④は成り立つ。
[Ⅱ] n=kのとき,④が成り立つ,すなわち
bk+1=bk
と仮定する。n=k+1のとき,②のn2kを代入して得られる等式と,2k−1を代入して得られる等式から
bk+2=
ck+k+1
k+1
ck+1=
k+ck
k+1

のなるので,bk+2
bk+2=
(k+k+1)k+1
bk+ck

と表される。したがって,⑤により,bk+2=bk+1が成り立つので,④はn=k+1のときにも成り立つ。
[Ⅰ],[Ⅱ]により,すべての自然数nに対して④の成り立つことが証明された。
 したがって,③が成り立つので,数列{bn}の一般項はbn=3である。
 次に,②のn2n−1に置き換えて得られる等式と③から
cn+1=13cn+1(n=1,2,3,)
となり,c1=であることと①から,数列{cn}の一般項は,(1)で求めた数列{pn}の一般項と等しくなることがわかる。

解説を読む

【2014年度センター試験.数学Ⅱ・B】第3問(選択問題)
 数列{an}の初項は6であり,{an}の階差数列は初項が9,公差が4の等差数列である。
(1) a2=アイa3=ウエである。数列{an}の一般項を求めよう。{an}の階差数列の第n項がn+であるから,数列{an}の一般項は
an=n+n+ ・・・①
である。

解説を読む

(2) 数列{bn}は,初項が25で,漸化式
bn+1=anan+11bn(n=1,,2,3) ・・・②
を満たすとする。b2=
シス
である。数列{bn}の一般項

と初項から第n項までの和Snを求めよう。
 ①,②により,すべての自然数nに対して
bn+1=
n+
n+
bn ・・・③

が成り立つことがわかる。

解説を読む

 ここで
cn=(n+)bn ・・・④
とするとき,③をcncn+1を用いて変形すると,すべての自然数nに対して
(n+)cn+1=(n+)cn
が成り立つことがわかる。これにより
dn=(n+)cn ・・・⑤
とおくと,すべての自然数nに対して,dn+1=dnが成り立つことがわかる。d1=であるから,すべての自然数nに対して,dn=である。したがって,④と⑤により,数列{bn}の一般項は
bn=
(n+)(n+)

である。また
bn=
n+
n+

が成り立つことを利用すると,数列{bn}の初項から第n項までの和Sn
Sn=
n
n+

であることがわかる。

解説を読む

【2015年度センター試験.数学Ⅱ・B】第3問(選択問題)
 自然数nに対し,2nの一の位の数をanとする。また、数列{bn}
b1=1,bn+1=anbn4(n=1,2,3,)・・・①
を満たすとする。
(1) a1=2, a2=a3=a4=a5=である。このことから,すべての自然数nに対して,a=anとなることがわかる。に当てはまるもの
を,次の⓪~④のうちから一つ選べ。
 ⓪ 5n ① 4n+1 ② n+3 ③ n+4 ④ n+5

解説を読む

(2) 数列{bn}の一般項を求めよう。①を繰り返し用いることにより
bn+4=
an+3an+2an+1an
2
(n=1, 2, 3, ・・・)

が成り立つことがわかる。ここで,an+3an+2an+1an=3·2
であることから,bn+4=
bnが成り立つ。このこと

から,自然数kに対して
b4k−3=(
)k−1
b4k−2=
(
)k−1

b4k−1=
(
)k−1
b4k=
(
)k−1

である。

解説を読む

(3) Sn=j=1nbjとおく。自然数mに対して
S4m=(
)m

である。
(4) 積b1b2···bnTnとおく。自然数kに対して
b4k−3b4k−2b4k−1b4k=
1
(
)(k−1)

であることから,自然数mに対して
T4m=
1
m
(
)m2m

である。また,T10を計算すると,T10=
3
2ヌネ
である。


解説を読む

【2016年度センター試験.数学Ⅱ・B】第3問(選択問題)
 真分数を分母の小さい順に,分母が同じ場合には分子の小さい順に並べてできる数列
12,13,23,14,24,34,15,
{an}とする。真分数とは,分子と分母がともに自然数で,分子が分母より小さい分数のことであり,上の数列では,約分できる形の分数も含めて並べている。以下の問題に分数形で解答する場合は,解答上の注意にあるように,それ以上約分できない形で答えよ。
(1) a15=
である。また,分母に初めて8が現れる

項は,aウエである。
(2) k2以上の自然数とする。数列{an}において,1kが初めて現れる項を第Mk項とし,k1kが初めて現れる項を第Nk項とすると
Mk=
k2
k+

Nk=
k2
k

である。よって,a104=
セソ
タチ
である。


解説を読む

(3) k2以上の自然数とする。数列{an}の第Mk項から
Nk項までの和は,
k−
である。したがって,

数列{an}の初項から第Nk項までの和は
k2
k

である。よって
n=1103an=
ハヒフ
ヘホ

である。

解説を読む

【2017年度センター試験.数学Ⅱ・B】第3問(選択問題)
 以下において考察する数列の項は,すべて実数であるとする。
(1) 等比数列{sn}の初項が1,公比が2であるとき
s1s2s3=s1+s2+s3=
である。
(2) {sn}を初項x,公比rの等比数列とする。a, bを実数(ただしa≠0)とし,{sn}の最初の3項が
s1s2s3=a3 ・・・①
s1+s2+s3=b ・・・②
を満たすとする。このとき
xr= ・・・③
である。さらに,②③を用いてr, a, bの満たす関係式を求めると
r2+()r+=0 ・・・④
を得る。④を満たす実数rが存在するので
a2+ab−b2≦0 ・・・⑤
である。
 逆に,a, bが⑤を満たすとき,③④を用いてr, xの値を求めることができる。

解説を読む

(3) a=64, b=336のとき,(2)の条件①,②を満たし,公比が1より大きい等比数列{sn}を考える。③,④を用いて{sn}の公比rと初項xを求めると,r=x=サシである。
 {sn}を用いて,数列{tn}
tn=snlogsn (n=1, 2, 3, ・・・)
と定める。このとき,{tn}の一般項は
tn=(n+n+である。{tn}の初項から第n項までの和Unは,UnUnを計算することにより
Uk=
n+
n+
テト

であることがわかる。

解説を読む

【2018年度センター試験.数学Ⅱ・B】第3問(選択問題)
 第4項が30,初項から第8項までの和が288である等差数列を{an}とし,{an}の初項から第n項までの和をSnとする。また,第2項が36,初項から第3項までの和が156である等比数列で公比が1よりも大きいものを{bn}とし,{bn}の初項から第n項までの和をTnとする。
(1) {an}の初項はアイ,公差はウエであり
Sn=n2カキn
である。
(2) {bn}の初項はクケ,公比はであり
Tn=(n)
である。

解説を読む

(3) 数列{cn}を次のように定義する。
ck=k=1n(nk+1)(akbk)
=n(a1b1)+(n1)(a2b2)++2(an1bn1)+(anbn)
(n=1, 2, 3, ・・・)
たとえば
c1=a1−b1,c2=2(a1−b1)+(a2−b2)
c3=3(a1−b1)+2(a2−b2)+(a3−b3)
である。数列{cn}の一般項を求めよう。
 {cn}の階差数列を{dn}とする。dn=cn+1−cnであるから,dn=を満たす。に当てはまるものを,次の⓪~⑦のうちから一つ選べ。
Sn+Tn
Sn−Tn
−Sn+Tn

−Sn−Tn
Sn+1+Tn+1
Sn+1−Tn+1

−Sn+1+Tn+1
−Sn+1−Tn+1

したがって,(1)と(2)により
dn=n2−2·n+
である。cn=ツテトであるから,{cn}の一般項は
cn=n3n2+n+n+
である。

解説を読む

【2019年度センター試験.数学Ⅱ・B】第3問(選択問題)
 初項が3,公比が4の等比数列の初項から第n項までの和をSnとする。また,数列{Tn}は,初項が−1であり,{Tn}の階差数列が数列{Sn}であるような数列とする。
(1) S2=アイT2=である。
(2) {Sn}{Tn}の一般項は,それぞれ
Sn=
Tn=
−n−

である。ただし,については,当てはまるものを,次の⓪~④のうちから一つずつ選べ。同じものを選んでもよい。
n−1 ① n ② n+1 ③ n+2 ④ n+3

解説を読む

(3) 数列{an}は,初項が−3であり,漸化式
nan+1=4(n+1)an+8Tn (n=1, 2, 3, ···)
を満たすとする。{an}の一般項を求めよう。
 そのために,bn=an+2Tnnにより定められる数列{bn}を考える。{bn}の初項はシスである。
 {Tn}は漸化式
Tn+1=Tn+n+ (n=1, 2, 3, ···)
を満たすから,{bn}は漸化式
bn+1=bn+ (n=1, 2, 3, ···)
を満たすことがわかる。よって,{bn}の一般項は
bn=テト
である。ただし,については,当てはまるものを,次の⓪~④のうちから一つ選べ。
n−1 ① n ② n+1 ③ n+2 ④ n+3
 したがって,{Tn}{bn}の一般項から{an}の一般項を求めると
an=
(n+)+

である。

解説を読む

【2020年度センター試験.数学Ⅱ・B】第3問(選択問題)
 数列{an}は,初項が0であり,n=1, 2, 3, ···のとき次の漸化式を満たすものとする。
an+1=n+3n+1{3an+3n+1(n+1)(n+2)} ・・・①
(1) a2=である。
(2) bn=an3n(n+1)(n+2)とおき,数列{bn}の一般項を求めよう。
 {bn}の初項b1である。①の両辺を3n+1(n+2)(n+3)で割ると
bn+1=bn+
(n+)(n+)
(
1
)n+1

を得る。ただし,<とする。
 したがって,
bn+1−bn=(
n+
n+
)(
1
)n+1

である。

解説を読む

 n2以上の自然数とするとき
k=1n1(
k+
k+
)=
1
(
n−
n+
)

k=1n1(
1
)k+1
=
(
1
)n

が成り立つことを利用すると
bn=
n−
(n+)
+
(
1
)n

が得られる。これはn=1のときも成り立つ。

解説を読む

(3) (2)により,{an}の一般項は
an=n−(n2)+
(n+)(n+)

で与えられる。ただし,<とする。
 このことから,すべての自然数nについて,anは整数となることがわかる。
(4) kを自然数とする。a3ka3k+1a3k+23で割った余りはそれぞれである。また,{an}の初項から第2020項までの和を3で割った余りはである。

解説を読む

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