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chủ đề 1: toán tính tổng theo quy luật Bai 1: Tính tổng 1000.999 1 4.3 1 3.2 1 2.1 1 ++++=S Giải: Ta có 1 11 )1()1( 1 )1( )1( )1( 1 + = + + + = + + = + kkkk k kk k kk kk kk (1) áp dụng đẳng thức (1) ta có: 1000 999 1000 1 1 1000 1 999 1 999 1 998 1 4 1 3 1 3 1 2 1 2 1 1 1000.999 1 4.3 1 3.2 1 2.1 1 1000 1 999 1 1000.999 1 999 1 998 1 999.998 1 . 4 1 3 1 4.3 1 3 1 2 1 3.2 1 2 1 1 1 2.1 1 == ++++=++++= = = = = = S S Bài 2: Tính tổng )1( 1 4.3 1 3.2 1 2.1 1 + ++++= nn S Giải: áp dụng đẳng thức (1) ta có: 11 1 1 1 11 4 1 3 1 3 1 2 1 2 1 1 )1.( 1 4.3 1 3.2 1 2.1 1 1 11 )1.( 1 . . 4 1 3 1 4.3 1 3 1 2 1 3.2 1 2 1 1 1 2.1 1 + = + = + ++++= + ++++= + = + = = = n n n S nnnn S nnnn Bµi 3: TÝnh tæng: 99.97 4 9.7 4 7.5 4 5,3 4 ++++=Q Gi¶i: Ta cã: 99 64 99 32 .2 99 133 .2 99 1 3 1 2 99 1 97 1 9 1 7 1 7 1 5 1 5 1 3 1 2 99.97 4 9.7 4 7.5 4 5,3 4 ______________________________________________________ 99 1 97 1 .2 99.97 4 9 1 7 1 .2 9.7 4 7 1 5 1 .2 7.5 4 5 1 3 1 .2 5.3 4 ==       − =       −=       −++−+−+−=++++=                        −=       −=       −=       −= Q Q Bµi 4: TÝnh tæng 100.97 1 11.8 1 8.5 1 5.2 1 ++++= S Gi¶i: Ta cã: 300 49 100 49 . 3 1 100 150 3 1 100 1 2 1 3 1 100 1 97 1 11 1 8 1 8 1 5 1 5 1 2 1 3 1 100.97 1 11.8 1 8.5 1 5.2 1 100 1 97 1 3 1 100.97 1 11 1 8 1 3 1 11.8 1 8 1 5 1 3 1 8.5 1 5 1 2 1 3 1 5.2 1 ==       − =       −=       −++−+−+−=++++= −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−                        −=       −=       −=       −= S S Bµi 5: TÝnh tæng )1.().1( 1 5.4.3 1 4.3.2 1 3.2.1 1 +− ++++= nnn S Gi¶i: Ta cã: )1.( 1 ).1( 1 )1.().1( 1 )1.().1( 1 )1.().1( )1()1( )1.().1( 2 + − − = +− + − +− + = +− −−+ = +− nnnnnnn n nnn n nnn nn nnn Suy ra: )1( )1( 1 ).1( 1 2 1 )1.().1( 1         + − − = +− nnnnnnn ¸p dông ®¼ng thøc (1) ta cã:         + ++++−         − ++++= −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−                          + − − = +−       −=       −=       −= )1( 1 5.4 1 4.3 1 3.2 1 2 1 ).1( 1 4.3 1 3.2 1 2.1 1 2 1 )1.( 1 .).1( 1 2 1 )1.().1( 1 5.4 1 4.3 1 2 1 5.4.3 1 4.3 1 3.2 1 2 1 4.3.2 1 3.2 1 2.1 1 2 1 3.2.1 1 nnnn S nnnnnnn Ta cã: n n n S nnnn S nnnn 11 1 1 1 1 4 1 3 1 3 1 2 1 2 1 1 )1.( 1 4.3 1 3.2 1 2.1 1 1 1 1 ).1( 1 . . 4 1 3 1 4.3 1 3 1 2 1 3.2 1 2 1 1 1 2.1 1 1 1 − =−= − − ++−+−+−= + ++++= −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−                − − = − −= −= −= 11 1 1 1 11 4 1 3 1 3 1 2 1 2 1 1 )1.( 1 4.3 1 3.2 1 2.1 1 1 11 )1.( 1 . . 4 1 3 1 4.3 1 3 1 2 1 3.2 1 2 1 1 1 2.1 1 2 2 + = + −= + −++−+−+−= + ++++= −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−                + −= + −= −= −= n n n S nnnn S nnnn VËy S = 2 1 (S 1 - S 2 )= )1(2 1 )1( 1 2 1 1 1 2 1 22 + −=         + −− =       + − − nnnn nn n n n n Bµi 6: TÝnh tæng )2)(1( 1 4,3,2 1 3.2.1 1 ++ +++= nnn S Gi¶i: Ta cã: )2)(1(2 1 )2)(1(2 122 )2)(1(2 )1()2( )2(2 1 )1(2 )2(2 1 2 1 . 2 1 )2)(1( 1 4.3 1 3.2 1 2 1 )1(21 . 2 1 )1( 1 3.2 1 2.1 1 2 1 )2)(1( 1 4.3 1 3.2 1 2 1 )1( 1 3.2 1 2.1 1 2 1 )2)(1( 1 )1( 1 2 1 )2)(1( 1 4.3 1 3.2 1 2 1 4.3.2 1 3.2 1 2.1 1 2 1 3.2.1 1 22 2 21 2 1 ++ −= ++ −−−+ = ++ +−+ = + + − + =−= + + = + + =         ++ +++= + = + =         + ++=         ++ +++−         + ++= −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−                    ++ − + = ++       −=       −= nnnn nnnn nn nnn n n n n SSS n n n n nn S n n n n nn S nnnn S nnnnnnn (2) Bµi 7: TÝnh tæng )3)(2)(1( 1 5.4.3.2 1 4.3.2.1 1 +++ +++= nnnn S Gi¶i: Ta có )1( )3)(2)(1( 1 )2)(1( 1 3 1 )3)(2)(1( )3( . 3 1 )3)(2)(1( 1 +++ ++ = +++ + = +++ kkkkkkkkkk kk kkkk áp dụng đẳng thức (1) ta có: )3)(2)(1(.3.4.3.2.1 4.3.2.1)3)(2)(1( )3)(2)(1( 1 4.3.2.1 1 3 1 )3)(2)(1( 1 5.4.3 1 4.3.2 1 4.3.2 1 3.2.1 1 3 1 )3)(2)(1( 1 5.4.3 1 4.3.2 1 3 1 5.4.3.2 1 4.3.2 1 3.2.1 1 3 1 4.3.2.1 1 +++ +++ = +++ = +++ ++= +++ = = nnnn nnnn nnnn S nnnnnnnn Bài 8: Dạng tổng theo quy luật S = a 1 + a 2 + a 3 + + a n Với d = a 2 a 1 = a 4 a 3 = =a n a n-1 Thì a n = a 1 + (n - 1).d 2 ).( 1 naa S n + = Bài 9: Tính tổng S = 1 + 2 + 3 + 4 + n Giải Cách 1:Ta có: S = 1+ 2+ 3+ 4+ +n + S = n + (n-1) + (n - 2) + (n - 3) + 1 2S = 1)(n 1) (n 1) (n ) 1(n ++++++++ = (n +1).n n lần 2 ).1( nn S + = Cách 2: Chọn hàm số g(x) = x Ta xác định hàm số f(x) bậc 2 có dạng f(x) = ax 2 + bx + c thoả mãn: g(x) = f(x) f(x -1) <=> x = ax 2 + bx + c a(x-1) 2 b(x-1) - c <=> x = ax 2 + bx + c ax 2 + 2ax - a bx + b - c <=> x = 2ax a + b = = = = 2 1 2 1 0 12 b a ab a cxxxf ++= 2 1 2 1 )( 2 (c tuỳ ý) Mặt khác: 2 ).1( 2 1 2 1 0. 2 1 0. 2 1 2 1 2 1 4321 )0()()( )4()3()2()1( )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( 222 nn nnccnnnS fnfnggggg nfnfng ffg ffg ffg ffg + =+= ++++=++++= =+++++ = = = = = Bài 10: Tính tổng: S = 1 + 3 + 5 + + (2n - 1) Giải: Đặt g(x) = 2x 1 Ta chọn hàm f(x) bậc 2 có dạng: f(x) = ax 2 + bx + c sao cho g(x) = f(x) f(x - 1) <=> 2x - 1 = ax 2 + bx + c a(x-1) 2 b(x-1) - c <=> 2x - 1 = ax 2 + bx + c ax 2 + 2ax - a bx + b - c <=> 2x 1 = 2ax a + b = = = = 0 1 1 22 b a ab a vậy f(x) = x 2 + c (c tuỳ ý) Ta có ( ) 2222 0)12 (531 )0()()( )4()3()2()1( )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( nccnccnnS fnfnggggg nfnfng ffg ffg ffg ffg =+=++=+++= =+++++ = = = = = Bài 11: Tính tổng S = 2 + 4 + 6 + + 2n Giải: Đặt g(x) = 2x Chọn hàm số f(x) có bậc 2 có dạng f(x) = ax 2 + bx +c sao cho: g(x) = f(x) f(x -1) <=> 2x = ax 2 + bx + c a(x-1) 2 b(x-1) - c <=> 2x = ax 2 + bx + c ax 2 + 2ax - a bx + b - c <=> 2x = 2ax a + b = = = = 1 1 0 22 b a ab a suy ra: f(x) = x 2 + x + c (c tuỳ ý) nnccnnnS fnfnggggg nfnfng ffg ffg ffg ffg +=++−++=++++= −=+++++ −−−−−−−−−−−−−−−−−−−−−−−−−−            −−= −= −= −= −= 222 )00(2 8642 )0()()( )4()3()2()1( )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( Bµi 12: TÝnh tæng S = 1 2 + 3 2 + 5 2 + + (2n - 1) 2 Gi¶i: §Æt g(x) = (2x -1) 2 Chän hµm sè f(x) cã bËc 3 cã d¹ng f(x) = ax 3 + bx 2 + cx + d sao cho: g(x) = f(x) – f(x-1) <=> (2x -1) 2 = ax 3 + bx 2 + cx + d – a(x-1) 3 – b(x-1) 2 – c(x-1) - d <=> 4x 2 – 4x + 1 = ax 3 + bx 2 + cx + d – ax 3 + 3ax 2 – 3ax + a – bx 2 + 2bx – b – cx + c - d <=> 4x 2 – 4x + 1 = 3ax 2 + (2b – 3a)x + a – b + c      =+− −=− = ⇔ 1 432 43 cba ab a        −= = = ⇔ 3 1 0 3 4 c b a nªn f(x) = dxx +− 3 1 3 4 3 (d tuú ý) 3 4 3 1 3 4 0. 3 1 0. 3 4 3 1 3 4 )12( 7531 )0()()( )4()3()2()1( )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( 3 33322222 nn nnddnnnS fnfnggggg nfnfng ffg ffg ffg ffg − =−=       +−−       +−=−+++++= −=+++++ −−−−−−−−−−−−−−−−−−−−−−−−−−            −−= −= −= −= −= Bµi 13: TÝnh tæng: S = 1 2 + 2 2 + 3 2 + n 2 Gi¶i: §Æt g(x) = x 2 Chän hµm sè f(x) cã bËc 3 cã d¹ng f(x) = ax 3 + bx 2 + cx + d sao cho: g(x) = f(x) – f(x-1) <=> x 2 = ax 3 + bx 2 + cx + d – a(x-1) 3 – b(x-1) 2 – c(x-1) - d <=> x 2 = ax 3 + bx 2 + cx + d – ax 3 + 3ax 2 – 3ax + a – bx 2 + 2bx – b – cx + c - d <=> x 2 = 3ax 2 + (2b – 3a)x + a – b + c          = = = ⇔      =+− =− = ⇔ 6 1 2 1 3 1 0 032 13 c b a cba ab a f(x)= dxxx +++ 6 1 2 1 3 1 23 (d tuú ý) 6 )12)(1( 6 32 6 1 2 1 3 1 0. 6 1 0. 2 1 0. 3 1 6 1 2 1 3 1 4321 )0()()( )4()3()2()1( )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( 23 23232322222 ++ = ++ = ++=       +++−       +++=+++++= −=+++++ −−−−−−−−−−−−−−−−−−−−−−−−−−            −−= −= −= −= −= nnnnnn S nnnddnnnnS fnfnggggg nfnfng ffg ffg ffg ffg Bµi 14: TÝnh tæng S = 2 2 + 4 2 + 6 2 + + (2n) 3 Gi¶i: §Æt g(x) = (2x) 2 Chän hµm sè f(x) cã bËc 3 cã d¹ng f(x) = ax 3 + bx 2 + cx + d sao cho: g(x) = f(x) – f(x-1) <=> 4x 2 = ax 3 + bx 2 + cx + d – a(x-1) 3 – b(x-1) 2 – c(x-1) - d <=> 4x 2 = ax 3 + bx 2 + cx + d – ax 3 + 3ax 2 – 3ax + a – bx 2 + 2bx – b – cx + c - d <=> 4x 2 = 3ax 2 + (2b – 3a)x + a – b + c        = = = ⇔      =+− =− = ⇔ 3 2 2 3 4 0 032 43 c b a cba ab a f(x)= dxxx +++ 3 2 2 3 4 23 (d tuú ý) 3 )132(2 3 264 3 2 2 3 4 0. 3 2 0.20. 3 4 3 2 2 3 4 )2( 8642 )0()()( )4()3()2()1( )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( 2 23 23232322222 ++ = ++ = ++=       +++−       +++=+++++= −=+++++ −−−−−−−−−−−−−−−−−−−−−−−−−−            −−= −= −= −= −= nnn nnn S nnnddnnnnS fnfnggggg nfnfng ffg ffg ffg ffg Bµi 15: TÝnh tæng: S = 1 3 + 2 3 + 3 3 + + n 3 Gi¶i: §Æt g(x) = x 3 Chän hµm sè f(x) cã bËc 4 cã d¹ng: f(x) = ax 4 + bx 3 + cx 2 + dx + e sao cho g(x) = f(x) – f(x-1) <=> x 3 = ax 4 + bx 3 + cx 2 + dx + e – a(x-1) 4 - b(x-1) 3 – c(x -1) 2 – d(x -1) – e <=> x 3 = ax 4 + bx 3 + cx 2 + dx + e – a(x 4 – 4x 3 + 6x 2 – 4x + 1) – b(x 3 – 3x 2 + 3x - 1) - c(x 2 – 2x + 1) – dx + d – e <=> x 3 = ax 4 + bx 3 + cx 2 + dx + e – ax 4 + 4ax 3 – 6ax 2 + 4ax – a – bx 3 + 3bx 2 – 3bx + b - cx 2 + 2cx – c – dx + d - e <=> x 3 = 4ax 3 + (3b – 6a)x 2 + (4a – 3b + 2c)x – a + b – c + d            = = = = ⇔        =+−+− =+− =− = ⇔ 0 4 1 2 1 4 1 0 0234 063 14 d c b a dcba cba ab a f(x) = exxx +++ 234 4 1 2 1 4 1 (e tuú ý) −−−−−−−−−−−−−−−−−−−−−−−−−−            −−= −= −= −= −= )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( nfnfng ffg ffg ffg ffg 2 2222 234 23423423433333 2 )1( 4 )1( 4 )12( 4 2 4 1 2 1 4 1 0. 4 1 0. 2 1 0. 4 1 4 1 2 1 4 1 4321 )0()()( )4()3()2()1(       + = + = ++ = ++ = ++=       +++−       +++=+++++= −=+++++ nnnnnnn nnn S nnneennnnS fnfnggggg Bµi 16: TÝnh tæng: S = 1 3 + 3 3 + 5 3 + + (2n - 1) 3 Gi¶i: §Æt g(x) = (2x - 1) 3 Chän f(x) bËc 3 cã d¹ng: f(x) = ax 4 + bx 3 + cx 2 + dx + e sao cho: g(x) = f(x) – f(x -1) <=> (2x -1) 3 = ax 4 + bx 3 + cx 2 + dx + e – a(x-1) 4 - b(x-1) 3 – c(x -1) 2 – d(x -1) – e <=> 8x 3 – 12x 2 + 6x -1 = ax 4 + bx 3 + cx 2 + dx + e – a(x 4 – 4x 3 + 6x 2 – 4x + 1) – b(x 3 – 3x 2 + 3x - 1) - c(x 2 – 2x + 1) – dx + d – e <=>8x 3 – 12x 2 + 6x -1 = ax 4 + bx 3 + cx 2 + dx + e – ax 4 + 4ax 3 – 6ax 2 + 4ax – a – bx 3 + 3bx 2 – 3bx + b - cx 2 + 2cx – c – dx + d - e <=> 8x 3 – 12x 2 + 6x -1 = 4ax 3 + (3b – 6a)x 2 + (4a – 3b + 2c)x – a + b – c + d        = −= = = ⇔        −=+−+− =+− −=− = ⇔ 0 1 0 2 1 6234 1263 84 d c b a dcba cba ab a f(x) = 2x 4 – x 2 + e (e tuú ý) −−−−−−−−−−−−−−−−−−−−−−−−−−            −−= −= −= −= −= )1()()( )3()4()4( )2()3()3( )1()2()2( )0()1()1( nfnfng ffg ffg ffg ffg ( ) ( ) 24 24242433333 2 200.22)12( 7531 )0()()( )4()3()2()1( nnS nneennnS fnfnggggg −= −=+−−+−=−+++++= −=+++++ Bµi 17: TÝnh tæng S = 2 3 + 4 3 + 6 3 + 8 3 + + (2n) 3 Gi¶i §Æt g(x) = (2x) 3 Chän f(x) bËc 3 cã d¹ng: f(x) = ax 4 + bx 3 + cx 2 + dx + e sao cho: g(x) = f(x) – f(x -1) <=> 8x 3 = ax 4 + bx 3 + cx 2 + dx + e – a(x-1) 4 - b(x-1) 3 – c(x -1) 2 – d(x -1) – e <=> 8x 3 = ax 4 + bx 3 + cx 2 + dx + e – a(x 4 – 4x 3 + 6x 2 – 4x + 1) – b(x 3 – 3x 2 + 3x - 1) - c(x 2 – 2x + 1) – dx + d – e <=>8x 3 = ax 4 + bx 3 + cx 2 + dx + e – ax 4 + 4ax 3 – 6ax 2 + 4ax – a – bx 3 + 3bx 2 – 3bx + b - cx 2 + 2cx – c – dx + d - e <=> 8x 3 = 4ax 3 + (3b – 6a)x 2 + (4a – 3b + 2c)x – a + b – c + d