漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト

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3 OK 3

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5 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1, 2 a n a n = 1+(n 1) 2 = 2n 1 a n+1 a n = b n a n+1 a n n a n+1 a n = (n ) 5

6 a n+1 a n = b n (n = 1, 2, 3, ) n 2 a n = a 1 + n 1 k=1 b k n 2 a n = a 1 + n 1 b k n = 1 a n = a 1 + n 1 b k k=1 k=1 n 2 S n S n 1 = a n a 1 a n+1 a n = b n n 2 n 1 k=1 n 1 1 n 2 n 1 (a k+1 a k ) = n 1 b k k=1 k=1 n 1 ( ) = n 1 (a k+1 a k1 ) k=1 = n 1 k=1 a k+1 n 1 a k k=1 = (a 2 + a 3 + a a n 1 + a n ) n 1 a k+1 (a 1 + a a n 2 + a n 2 ) n 1 k=1 k=1 a k = a n a 1 6

7 a n a 1 = n 1 b k a n = a 1 + n 1 b k k=1 k=1 a n+1 a n = 2n n 2 a n = 3 + n 1 k=1 2k a n = a 1 + n 1 = (n 1){(n 1) + 1} 2 = 3 + n(n 1) = n 2 n + 3 k=1 b k a n = n 2 n + 3 n = 1 a 1 = = 3 n = 1 a n = n 2 n + 3 a n = n 2 n + 3 (n 1) 7

8 {a n } (1) a 1 = 3, a n+1 = 3a n (2) a 1 = 4, a n+1 3 = 2(a n 3) a n+1 = 3a n a n+1 a n a n+1 a n 3 {a n } 3 3 a n a n = 3 3 n 1 = 3 n a n 3 = b n b n+1 = a n+1 3 b n b n+1 b n n n + 1 b n+1 b n = a n 3 b n+1 = a n+1 3 a n+1 3 = 2(a n 3) b n+1 = 2b n b n 3 a n 3 = b n b 1 = a 1 3 = 4 3 = 1 b n {b n } 1 2 b n = 1 2 n 1 8

9 a n 3 = 2 n 1 a n = 2 n a n 3 = b n {a n 3} a 1 3 = 4 3 = 1 2 a n 3 = 1 2 n 1 a n = 2 n

10 {a n } (1) a 1 = 2, a n+1 = 3a n 2 (2) a 1 = 1, 2a n+1 = a n 2 a n+1 = Aa n + B a n+1 = 3a n a n+1 α = β(a n α) 2 {a n α} β α β 2 a n+1 = βa n αβ + α 1 a n 3 = β, 2 = αβ + α α = 1, β = 3 α = 1, β = 3 2 a n+1 1 = 3(a n 1) a n+1 α = β (a n α) 2 β = 3 a n+1 = 3a n 2 1 β = 3 α α a n+1 = Aa n + B a n+1 α = A (a n α) α α = Aα + B α = Aα + B 10

11 α α a n+1 = Aa n + B ) α = Aα + B a n+1 α = Aa n Aα = A(a n α) a n+1 = Aa n + B α = Aα + B α a n+1 α = A (a n α) a n+1 a n a n+1 a n α α a n+1 = Aa n + B a n+1 = Aa n + B a n+1 = Aa n + B 1 α = Aα B α 1 a n+1 α = A(a n α) 11

12 a n+1 = 3a n 2 1 α =3α 2 α =1 1 a n+1 1 = 3(a n 1) {a n 1} a 1 1 = 2 1 = 1 3 a n 1 = 1 3 n 1 a n = 3 n a n+1 = Aa n + B a n+1 1 a n+1 1 a n+1 = Aa n + B 2a n+1 = a n 2 a n+1 = 1 2 a n a n+1 = Aa n + B α = 1 α 1 α = a n+1 ( 2) = 1 2 {a n ( 2)} a n = 1 2 (a n + 2) {a n + 2} a = = ( ) a n + 2 = 3 1 n 1 2 ( ) 1 n 1 a n =

13 a 1 = 4, a n+1 = 6a n + 2 n+2 {a n } a n+1 = Aa n + B C n a n+1 = 6a n + 2 n+2 a n = Aa n + B C n a n+1 = 6a n n a n+1 = Aa n + B C n a n+1 = Aa n + B C n a n+1 = Aa n + B C n C n+1 b n = a n C n a n+1 = Aa n + B ( a n+1 a n = ( ) ) a n+1 = Aa n + B C n A n+1 b n = a n A n 13

14 a n+1 = 6a n + 2 n+2 a n+1 = 6a n n a n+1 2 = 6a n + 4 2n 2 n+1 n+1 n n+1 2 n n 6a n 2 n+1 = 6a n 2 2 n = 3a n 2 n = 3 an n 4 2n = 4 2n 2 2 n n = 2 a n+1 an = 3 n n + 2 b n = a n 2 n b 1 = a = 4 2 = 2 b n+1 = 3b n a n+1 = Aa n + B a n+1 = Aa n + B α = 3α + 2 α = 1 1 b n = 3(b n + 1) {b n + 1} b = = 3 3 b n + 1 = 3 3 n 1 = 3 n b n = 3 n 1 b n = a n 2 n a n 2 n = 3n 1 b n = a n 2 n a n = a n 2 n a n = 2 n (3 n 1) 2 n = 6 n 2 n 2 n 3 n = (2 3) n = 6 n (ab) 2 = a n b n 14

15 a n+1 = 6a n + 2 n+2 a n+1 = 6a n n a n+1 6 = 6a n + 4 2n 6 n+1 n+1 n n+1 6 n n 6a n 6 = 6a n n n = a n 6 n 4 2n = 4 2n 6 n n = 4 ( ) 2 n ( ) = 2 1 n a n+1 6 = a n n+1 6 n + 2 ( ) 1 n 3 3 b n = a n 6 n b 1 = a = 4 6 = 2 3 b n = b n + 2 ( ) 1 n 3 3 b n+1 b n = 2 ( ) 1 n 3 3 n 2 b n = b 1 + n 1 k=1 = = ( ) 1 n 3 { ( ) n 1 } n 1 ( ) 1 n 1 k= n { ( ) 1 1 n 1 } = { ( ) 1 1 n 1 } 3 3 ( ) = 1 1 n 3 ( ) n = 1 n 1 b n = 1 1 n 3 n = 1 b n = a n 6 n a ( ) n 6 n = 1 1 n a 3 n = 6 n 2 n 15

16 a 1 = 1 2, a n+1 = a n 2a n + 3 {a n} a n+1 = Aa n Ba n + C 0 0 a n+1 = Aa n Ba n +C b n = 1 a n a n+1 = Aa n + B a a n+1 = n 2a n a n \= 0 a n \= 0 a n+1 = 0 1 a n = 0 a n = a n 1 = = a 1 = 0 a 1 = 1 2 n a n \= 0 a n \= 0 a n \= 0, a n+1 \=

17 1 = 2a n + 3 a n+1 a n 1 = 2a n + 3 a n+1 a n a n 1 = a n+1 a n 1 a n = b n b n+1 = 3b n b 1 = 1 a 1 = = = = 2 1 = 2 α = 3α + 2 α = 1 2 b n = 3(b n + 1) {b n + 1} b = 3 3 b n + 1 = 3 3 n 1 b n = 3 n 1 b n = 1 a n a n = 1 b n = 1 3 n 1 a n = 1 3 n 1 17

18 a 1 = 1, a n+1 = 2a n + n 1 {a n } a n+1 = Aa n + Bn + C a n+1 a n a n+1 + α(n + 1) + β = A(a n + αn + β) a n+1 = Aa n + Bn + C 1 a n+1 = Aa n + Bn + C 1 n n + 1 a n+2 = Aa n+1 + B(n + 1) + C a n+2 = Aa n+1 + B(n + 1) + C ) a n+1 = Aa n + Bn + C a n+2 a n+1 = A(a n+1 a n ) + B b n = a n+1 a n a n+2 a n+1 = A(a n+1 a n ) + B b n+1 = Ab n + B b n a n a n+1 = 2a n + n a n+2 = 2a n+1 + n

19 a n+2 a n+1 = 2(a n+1 a n ) + 1 b n = a n+1 a n b n+1 = 2b n n = 1 a 2 = 2a 1 1 = = 2 b 1 = a 2 a 1 = 2 1 = 1 {b n } {b n } b 1 = 1, b n+1 = 2b n α = 2α + 1 α = 1 3 b n = 2(b n + 1) {b n + 1} b = = 2 2 b n + 1 = 2 2 n 1 = 2 n b n = 2 n 1 a n+1 a n = b n a n+1 a n = 2 n 1 n 2 a n = a 1 + n 1 (2 k 1) k=1 = 1 + 2(2n 1 1) (n 1) 2 1 = 1 + 2(2 n 1 1) n + 1 = 2 n n a n = 2 n n n = 1 a n = 2 n n 19

20 a n+1 = Aa n + Bn + C 1 a n+1 = Aa n + Bn + C 1 a n+1 + α(n + 1) + β = A(a n + αn + β) α, β 2 {a n } + αn + β a 1 + α + β A 2 a n+1 + α(n + 1) + β = A(a n + αn + β) n + 1 n n + 1 n a n+1 α = A(a n α) n + 1 n n + 1 n 20

21 a n+1 = 2a n + n 1 1 α, β a n+1 + α(n + 1) + β = 2(a n + αn + β) 2 21

22 2 a n+1 = 2a n + αn + β α 1 α = 1, β α = 1 α = 1, β = 0 2 a n+1 + (n + 1) = 2(a n + n) {a n + n} a = 2 2 a n + n = 2 2 n 1 a n = 2 n n OK 22

23 {a n } n S n S n = n 2 + 4n {a n } S n S n 1 S 1 = a 1 2 n 2 a n = S n S n 1 1 S 1 = a 1 S n n S 1 S 1 = a 1 S n a 1 S 1 = a 1 a 1 2 n 2 a n = S n S n 1 S n n S n S n = a 1 + a a n 1 + a n S n 1 n 1 S n 1 = a 1 + a a n 1 S n S n 1 S n = a 1 + a a n 1 + a n ) S n 1 = a 1 + a a n 1 S n S n 1 = a n 23

24 S n S n 1 S n S n 1 = a n S n = 1 S n 1 n = 1 S 1 1 = S 0 S 0 0 S n S n 1 = a n n 2 S n 2 n 2 a n n = 1 n 2 a n = a 1 + n 1 k=1 b k a n a n n = 1 S n = n 2 + 4n a 1 = S 1 = = 5 n 2 a n = S n S n 1 = n 2 + 4n { (n 1) 2 + 4(n 1) } = n 2 + 4n n 2 + 2n 1 4n + 4 = 2n + 3 n = 1 a n = 2n + 3 n = 1 a 1 = = 5 n = 1 a n = 2n + 3 a n = 2n + 3 (n 1) 24

25 {a n } (1) a 1 = 1, a 2 = 5, a n+2 5a n+1 + 6a n = 0 (2) a 1 = 1, a 2 = 2, a n+2 + 3a n+1 4a n = 0 (3) a 1 = 1, a 2 = 5, a n+2 6a n+1 + 9a n = 0 a n+2 + Aa n+1 + Ba n = 0 a n+2 + Aa n+1 + Ba n = 0 a n+2 + Ba n+1 + Ca n = 0 1 x 2 + Bx + C = 0 α, β 1 a n+2 αa n+1 = β (a n+1 αa n ) 2 a n+2 βa n+1 = α (a n+1 βa n ) 3 2 {a n+1 αa n } β 3 {a n+1 βa n } α α 1 a n+2 αa n+1 = α (a n+1 αa n ) 25

26 x 2 + Bx + C = 0 α, β α + β = B, αβ = C 2, 3 a n+2 (α+β)a n+1 +αβa n = 0 α + β = B, αβ = C a n+2 + Ba n+1 + Ca n = x 2 5x + 6 = a n+1 αa n a n+1 βa n 26

27 a 1 = 1, a 2 = 5, a n+2 5a n+1 + 6a n = 0 1 x 2 5x + 6 = 0 x = 2, 3 1 a n+2 2a n+1 = 3 (a n+1 2a n ) 2 a n+2 3a n+1 = 2(a n+1 2a n ) 3 2, 3 2 {a n+1 2a n } a 2 2a 1 = = 3 3 a n+1 2a n = 3 3 n 1 = 3 n 2 3 {a n+1 3a n } a 2 3a 1 = = 2 2 a n+1 3a n = 2 2 n 1 = 2 n a n+1 a n 2 3 a n+1 2a n = 3 n ) a n+1 3a n = 2 n a n = 3 n 2 n a n = 3 n 2 n 2 3 a n+1 = Aa n + B C n

28 2 a n+1 2a n = 3 3 n 1 = 3 n a n+1 = 2a n + 3 n a n+1 3 n+1 = 2 3 an 3 n n+1 b n = a n 3 n b 1 = a 1 3 = 1 3 b n+1 = 2 3 b n α = 2 3 α α = 1 2 b n+1 1 = 2 3 (b n 1) {b n 1} b 1 1 = = b n 1 = 2 ( ) 3 2 n 1 ( ) = 2 n 3 3 b n = a n 3 n a ( ) n 3 n = 2 n a n = 3 n 2 n x 2 + 3x 4 = 0 x = 1, 4 28

29 1 a n+1 a n a 1 = 1, a 2 = 2, a n+2 + 3a n+1 4a n = 0 1 x 2 + 3x 4 = 0 x = 1, 4 1 a n+2 a n+1 = 4(a n+1 a 1 ) {a n+1 a n } a 2 a 1 = 2 1 = 1 4 a n+1 a n = ( 4) n 1 n 2 a n = 1 + n 1 ( 4) k 1 k=1 = ( 4)n 1 1 ( 4) = 6 ( 4)n 1 5 a n = 6 ( 4)n 1 5 n = 1 a n = 6 ( 4)n 1 5 a n+2 αa n+1 = β(a n+1 a n ) α = 4, β = 1 a n+2 + 4a n+1 = a n+1 + 4a n a 1 = 1, a 2 = 2, a n+2 + 3a n+1 4a n = 0 1 x 2 + 3x 4 = 0 x = 1, 4 29

30 1 a n+2 + 4a n+1 = a n+1 + 4a n {a n+1 + 4a n } a n+1 + 4a n = a n + 4a n 1 = = a 2 + 4a 1 = = 6 b n = a n+1 + 4a n b n+1 = a n+2 + 4a n+1 a n+2 + 4a n+1 = a n+1 + 4a n b n+1 = b n b n = b 1 a n+1 + 4a n = a 2 + a 1 = 6 a n+1 +4a n = 6 a n++1 = 4a n +6 1 a n+1 = Aa n + B α = 4α + 6 α = a n+1 6 ( 5 = 4 a n 6 ) 5 { a n 6 } a = = a n 6 5 = 1 5 ( 4)n 1 a n = 6 ( 4)n 1 5 a n+2 6a n+1 + 9a n = 0 1 x 2 6x + 9 = 0 x = 3 30

31 1 a n+2 3a n+1 = 3 (a n+1 3a n ) {a n+1 3a n } a 2 3a 1 = = 2 3 a n+1 3a n = 2 3 n 1 a n+1 = 3a n n 1 a n+1 3 n+1 = a n 3 n n+1 b n = a n 3 n b n+1 = b n b n+1 b n = 2 9 b n b 1 = a = b n = 1 3 b n = a n 3 n a n 3 n = 2n (n 1) 2 9 = 2n a n = 3 n 2 (2n + 1) 31

32 a 1 = 1, b 1 = 3, a n+1 = 2a n + b n, b n+1 = a n + 2b n {a n } {b n } {a n } {b n } a n+1 = 2a n + b n 1, b n+1 = a n + 2b n , , a n+1 + b n+1 = 2a n + b n + a n + 2b n = 3 (a n + b n ) {a n + b n } 1 2 a n+1 b n+1 = 2a n + b n a n 2b n = a n b n {a n b n } 32

33 1 + 2 a n+1 + b n+1 = 2a n + b n + a n + 2b n = 3 (a n + b n ) {a n + b n } a 1 + b 1 = = 4 3 a n + b n = 4 3 n a n+1 + b n+1 = 2a n + b n a n 2b n = a n b n {a n + b n } a 1 b 1 = 1 3 = 2 a n b n = a n + b n = 4 3 n 1 +) a n b n = 2 2a n = 4 3 n 1 2 a n + b n = 4 3 n 1 +) a n b n = 2 2b n = 4 3 n a n = 2 3 n 1 1 b n = 2 3 n

34 a 1 = 1, b 1 = 3, a n+1 = 3a n + b n, b n+1 = 2a n + 4b n {a n } {b n } {a n } {b n } III a n b n a n+1 a n b n+1 b n a n+1 + αb n+1 = β (a n + αb n ) α, β {a n + αb n } β a n+1 = 3a n + b n 1 b n+1 = 2a n + 4b n 2 1 b n = a n+1 3a n 1 b n+1 = a n+2 3a n n n

35 1, 1 2 a n+2 3a n+1 = 2a n + 4(a n+1 3a n ) a n+2 7a n a n = 0 3 x 2 7x + 10 = 0 (x 2)(x 5) = 0 3 a n+2 2a n+1 = 5 (a n+1 2a n ) {a n+1 2a n } a 2 2a 1 = (3a 1 +b 1 ) 2a 1 1 a 2 = 3a 1 + b 1 = a 1 +b 1 = 4 5 a n+1 2a n = 4 5 n a n+2 5a n+1 = 2 (a n+1 5a n ) {a n+1 5a n } a 2 5a 1 = (3a 1 + b 1 ) 5a 1 = 2a 1 + b 1 = = 1 2 a n+1 5a n = 2 n a n+1 2a n (a n+1 5a n ) = 4 5 n 1 2 n 1 3a n = 4 5 n 1 2 n 1 a n = 4 3 5n n n n + 1 a n+1 = 4 3 5(n+1) (n+1) 1 = n n 1 6 6, 6 1 b n = n 1 2 ( 3 2n n 1 1 ) 3 2n 1 = 8 3 5n n 1 a n = 4 3 5n n 1, b n = 8 3 5n n 1 35

36 a n+1 + βb n+1 = α(a n + βb n ) a n+1 + βb n+1 = α(a n + βb n ) α, β a n+1 + βb n+1 = α(a n + βb n ) {a n + βb n } α a n+1 = 3a n + b n 1 b n+1 = 2a n + 4b n 2 α, β a n+1 + βb n+1 = α(a n + βb n ) = αa n + αβb n 3 α, β a n+1 + βb n+1 = 3a n + b n + β(2a n + 4b n ) ( 1, 2 ) = (3 + 2β) a n + (1 + 4β) b n 4 3, β = α 5, 1 + 4β = αβ 6 a n b n 5, 6 (α, β) = ( 2. 1 ), (5, 1) 2 (α, β) = ( 2. 1 ) 3 α = 2, β = a n+1 1 ( 2 b n = 2 a n 1 ) 2 b n { a n 1 } 2 b n a b 1 = = a n 1 2 b n = 1 2 2n

37 (α, β) = (5.1) 3 α = 5, β = 1 a n+1 + b n = 5 (a n + b n ) {a n + b n } a 1 + b 1 = = 4 5 a n + b n = 4 5 n a n 1 2 b n (a n + b n ) = 1 2 2n n b n = 1 2 2n n 1 b n = 1 3 2n n a n n n 1 = 4 5 n 1 a n = 4 5 n n n 1 = 4 3 5n n 1 a n = 4 3 5n n 1, b n = 8 3 5n n 1 37

38 {a n } (1) a 1 = 1, (n + 2)a n+1 = na n (2) a 1 = 5, a n+1 = 2n 1 2n + 3 a n (3) a 1 = 1, (n + 3)a n+1 = 2na n (n + 2)a n+1 = na n n + 2 n n + 1 (n + 2)a n+1 = na n n + 1 (n + 2)(n + 1)a n+1 = (n + 1) n a n b n = (n + 1) n a n b n+1 = (n + 2)(n + 1)a n+1 (n + 2)(n + 1)a n+1 = (n + 1) n a n b n+1 = b n {b n } b n = b 1 (n + 2)a n+1 = na n (n + 1) (n + 2)(n + 1)a n+1 = (n + 1) n a n {(n + 1)na n } (n + 1)na n = (1 + 1) 1 a 1 = 2 38

39 a n = 2 n (n + 1) a n+1 = 2n 1 2n + 3 a n 2n + 3 2n + 3 n + 1 n 2n 1 2n + 3 2n + 3 a n = 2n + 3 2n 1 a n 2n + 3 a n 2n + 3 = a n 2n 1 2n + 3 2n + 3 (2n + 3)a n+1 = (2n 1)a n 2n + 3 2n 1 2n + 1 (2n + 3)(2n + 1)a n+1 = (2n + 1)(2n 1)a n b n = (2n + 1)(2n 1)a n b n+1 = b n {b n } 39

40 a n+1 = 2n 1 2n + 3 a n (2n + 3)a n+1 = (2n 1)a n (2n + 3)(2n + 1)a n+1 = (2n + 1)(2n 1)a n (2n + 1) {(2n + 1)(2n 1)a n } (2n + 1)(2n 1)a n = ( )(2 1 1) 5 = 15 a n = 15 (2n + 1)(2n 1) (n + 3)a n+1 = 2na n n + 3 n n n (n + 1)(n + 2) n n + 3 n + 1 n + 2 (n + 1)(n + 2) 40

41 41

42 (n + 3)a n+1 = 2na n (n + 2)(n + 1) (n + 3)(n + 2)(n + 1)a n+1 = 2(n + 2)(n + 1)na n b n = (n + 2)(n + 1)na n b n+1 = 2b n b 1 = (1 + 2) (1 + 1) 1 a 1 = 6 {b n } 6 2 b n = 6 2 n 1 = 3 2 n (n + 2)(n + 1)na n = 3 2 n a n = 3 2 n n(n + 1)(n + 2) 42

43 a 1 = 1, a n+1 = a 2 n + 2n a n 2 (n = 1, 2, 3, ) {a n } (1) a 2, a 3, a 4 (2) a n a 2, a 3, a 4 {a n } a n+1 = a 2 n + 2n a n 2 a n a 2, a 3, a 4 a n+1 = a 2 n + 2n a n 2 n = 1, 2, 3 a 2 = a a 1 2 a n+1 = a 2 n + 2n a n 2 n = 1 = ( 1) ( 1) 2 = = 3 43

44 a 3 = a a 2 2 a n+1 = a 2 n + 2n a n 2 n = 2 = ( 3) ( 3) 2 = = 5 a 4 = a a 3 2 a n+1 = a 2 n + 2n a n 2 n = 3 = ( 5) ( 5) 2 = = 7 a 1 = 1, a 2 = 3, a 3 = 5, a 4 = a n = 1 + (n 1) ( 2) = 2n + 1 a 1, a 2, a 3, a 4 a n = 2n + 1 1, 2, 3, 4 n 5 a n = 2n + 1 a n = 2n + 1 a n = 2n n = 1 1 n = k 1 n = k + 1 n = 1 OK 44

45 n = k 1 n = k n = k 1 a k = 2k + 1 n = k a k+1 = 2(k + 1) + 1 OK a n+1 = a 2 n + 2n a n 2 n k a k+1 n k a n+1 = a 2 n + 2n a n 2 n n n k OK k a k = k + 1 a k+1 = 2(k + 1) + 1 a n = 2n + 1 a n = 2n (i) n = 1 a 1 = 1 1 (ii) n = k (k ) 1 a k = 2k

46 a n+1 = a 2 n + 2n a n 2 n k a k+1 = a 2 k + 2k a k 2 = ( 2k + 1) 2 + 2k ( 2k + 1) 2 ( a k = 2k + 1) = 4k 2 4k + 1 4k 2 + 2k 2 = 2k 1 = 2(k + 1) n = k + 1 n a n = 2n + 1 a n = 2n

47 a 1 = 1, a n+1 = 2a n + 1 3a n + 4 {a n} a n+1 = Ca n + D Aa n + B Aa a n+1 = n Ba n + C Ca n + D a n+1 = Ca n + D Aa n + B 1 x = Cx + D Ax + B α, β α, β 1 α, β 47

48 a n+1 = 2a n + 1 3a n x = 2x + 1 3x + 4 x = 1, a n+1 ( 1) = 2a n + 1 3a n + 4 ( 1) a n = 5(a n + 1) 3a n x = 1 x = a n = 2a n + 1 3a n = 3a n 1 3(3a n + 4) = a n 1 3 3a n

49 ( a n \= 1) a 1 = 1 > 0 a n+1 = 2a n + 1 3a n + 4 n a n > 0 a n + 1 \= 0 a n > 0 a n + 1 \= 0 a n \= 1 a n > 0 a n \= 1 a n+1 = 2a n + 1 3a n + 4 a n > 0 a n+1 > 0 a n > 0 a n \= 1 a n > 0 a n > 0 a n > 0 a n \= , 3 a n a n = a n 1 3 3a n + 4 5(a n + 1) (3a n + 4) = 1 5 a n 1 3 a n

50 a n a n = 1 5 a n 1 3 a n + 1 a n 1 3 a n + 1 a n 1 3 a n + 1 a a = = a n 1 3 a n + 1 = 1 ( ) 3 1 n 1 5 a n a n a n 1 3 a n + 1 = 1 ( ) 3 1 n 1 5 = ( n n 1 a n 1 ) = a 3 n n 1 a n 5 n 1 = a n + 1 (3 5 n 1 1)a n = 5 n a n = 5n n

51 51

52

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3 13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >

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