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b x( n1xn)a x( nxn1) であり， 0 b より n 1 n a ( n n 1) x x x x

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シェア "b x( n1xn)a x( nxn1) であり， 0 b より n 1 n a ( n n 1) x x x x"

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[ 東京工業大学 1959 年 解析Ⅱ １ ]

0 a 1

b のとき，次の条件を満たす数列{ }xn の一般項を求めよ。

1 1 ( ) 0

n n n

ax bx  a b x 0 1

0

( 1), 0, n 1

n

n ax bx x

≧

1 1 ( ) 0

n n n

ax bx  a b x ⇔ b x( n1xn)a x( nxn1) であり，

0

b より n 1 n a ( n n 1)

x x x x

b

2

1 2

( n n )

a x x

b

( 1 0) a n

x x b

となる。

0 1 0

ax bx

より 1 a 0

x x

b から 1 0 0

n

n n

a a

x x x x

b b

    1 0

a n a b b x

 

  

n≧1 のとき

1

0 0

1

1

n k n

k

a a

x x x

b b

 

  

1

0 0

0

1

n k

k

a a

x x

b b

0 0

1 1

1 a n

a b

x x

b a

b

  

0

a n

b x

   これは n0 のときも成り立っている。

さらに，

0 n 1

n

x

と 0 ba 1 より

0 0

0

1 1

n

n

x a x

b a

b

から x0  1 ab

よって 1

n n

a a

x b b

 

   となる。

参照

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