JtyvaHpC!
Mot phifofng phap tim lofi giai cho cac bai toan tfnh tdng day so'viet theo quy luat
I. DAT VAN DE
Trong chuang trinh sd hpc d trudng THCS cd dang toin
"Diy s6 vi^t theo quy lu|t" li d^ing toin tuong ddi khd ddi vdi hpc sinh, tdng hpp nhi^u kid thiic, cac em phii phan tfch, phin doin, nh§n dang bii toin dk dua ra quy luSt cua diy so, dua ra hudng giii. Cd the nh^n thay hpc sinh thudng IQng tiing, khd khin trong vi?c tim hudng giai, khd hieu khi diing trude dang toin niy. VI vay tdi manh dan chpn chuyen dk "Mpt phinmg phip tim Idi giii cho cic bii toin tfnh tdng d3y s6 vi^t theo quy luat".
Rit nhi^u sich tham khio viet vk van di niy, hpc sinh khi dpc sich thi ed the hieu dupe Idi giii nhung khdng hilu tai sao Igi ed the tim ra Idi giii nhu vay. ChuySn d^ niy trinh biy cich hudng din hpe sinh tiSp c§n cic bii toin tinh tdng ciia diy so vi^t theo quy luit theo hudng ty nhien, gde rl cua vin dk.
II. N(>1 DUNG
1. N$i dung chfnh cua sing kien
Xin dua ra mpt bii toin khi quen thupe cho hpc sinh khi, gidi trong chuong trinh s6 hpc Idp 6 nhu sau.
1.1. Bii toin 1
Tinh ting S- 1.2 + 2.3 + 3.4 +... + 99.100 Vdi bii toin niy cich giii mi cic sich tham khio dua ra nhu sau:
TirS= 1.2 + 2.3 + 3.4 + ... + 99.100 suyra3.S= 1.2.3+ 2.3.3+ 3.4.3 + ... + 99.100.3
= 1.2.3+2.3.(4-l)+3.4.(5-2)+...+99.100.(101-98)
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 + ...+99.100.I01-98.99.100
= 99.100.101
Dodd 5=99.100.101:3 = 333300
Ldi giii trln rit ngin gpn, hpe sinh cung rat dl tiep thu, hi^u bii. Nhung chfle ft hpc sinh ty djt cho minh eSu hdi li tgi sao la Igi biit nhin ei 2 vi ding thQc ban dau vdi 3.
Vdi cic tdng khic (ed thi phde tgp Hon) thi cd cdn lim nhu v^y dupe ntia hay khdng, cd phuang phip nio chung cho cic bii toin dgng tr£n hay khdng.
D^ (ri Idi dupc cflu hdi tr£n, tdi xin dua ra cich hudng din hpc sinh tiip c$n bii toin nhu sau:
Vdi tdng 3=1.2 + 2.3 + 3.4 + ...+ 99.100 thi cic sd hgng diu ed dgng n(n+l)=n^ + n vdi H G N ' , n <99.
V^y niu ta tim dupc
fi;x)mithdaman f(n+l)-fl;n)= n ' + n v d i n e N * .
HOANG VAN LONG TrUdng THCS LS Qu;? Ddn - TP. H i i DUdng Thi ta cd dupe;
S=f(2>f(l) + f(3)-fl:2) + fl:4)-fl:3)+....+ f(100)-fl;99)
=f(100)-f(l)
Vay miu chdt cua bai toin tr6n la tim ra f(x).
Vdi eieh giai trong eie sich tham khao ta thiy 2.3.3=2.3 4-1.2.3
3.4.3-3.4.5-2.3.4 4.5.3=4.5.6-3.4.5 99.100.3=99.100.101-98.99.100 Vay thyc ra li nhin tdng S vdi 3 de cd
3.S- 1.2.3 + 2.3.3 + 3.4.3 +... + 99.100.3 Tir dd mSi so hang cd thi viit dupe thinh hi^u d^ng f(n)- fln-l).
VSy vdi moi bii khie nhau thi cic sl cin nhin thfim khic nhau. Thim tri doi vdi cic bii toin phUc tap dgng niy, hpc sinh vi ngay d giio vien eiJng thiy IQng tQng khi tim ldi giii.
Xin dua ra eich hudng dSn cic em tim ra f(x), mau ch6t cho vi$c giai cic bii toin dang tr€n.
Phin tich;
Ta nh^ thay sd hang tdng quit ciia diy n(n+l)=n^ + n 1^
bac 2 nen de cd
fl;n+l)-fl;n>= n(n+l)=n^ + n thi ta phii tim da thQc f^x) bac 3 di fl:x+l)-fl;x)= x^+x Vx .
Tuy nhign chi cin x^t da thuc dang
i{xy= a.x^+b.x^+c.x vi h? sd tu do cQa f{x+l) vi f(x) li nhu nhau.
Vi^c xie dinh da thUe f(x) dua tren vio vigc d6ng nhat h$
sdhaidathucfl:x+l)-f(x)= X^+X Vx
^ a(x+iy-43.{x+l)^+c.(x+l>{ax'+b.x^-K;.x)^^+x W
^ 3a.x^ + 3ax+2b.x + a+b+c = x^ +x Vx
3a = 1 3a+2b = 1 4 a+b+c = 0
a = -I 3 b = 0 =
_ ^ ] _
3
> f ( x ) = - x - - X 3 3
V$yS=ftlOO)-fl:i)= = - , 1 0 0 ' - . ! 0 0 - - . l ' - . l =333300
48 M S mssms'
Chu y trong khi trmh My Icri giai cho bai toan, di co Itri giai ng3n gon ta CO thg b6 qua buoc tim da thiic %y^ (Chj thuc hien buoc nay o gidy iihap). Sau do xet ngay da thuc f(x) chi ra rang S;x+l)-«;x)= X^+X V x . VI dn v4 each trinh b4y ]6i giai cho bai toan:
1.2. Bii toin 2
Rutgpntdng S=l-2+2J+3.4+...+n(iH-l) vdi n e N * . Hudng din
X« da thuc f(x)- - x ' — x khi d6 ta ci: fi;x+l>fi;x)- -!-(x+l)'-i(x+l) - -i-x'-ix = x ' + x = x ( x + l ) Vx
|3 3 I [3 3 J Ap dung cho X bang 1; 2; 3;...; n taduoc S-1.2 + 2.3 + 3.4 + .., + n(n+l)
=fi:2>«:i)+fl;3>f(2)+f(4)-fi:3)+...+s:n+i>fl:n) - fi;n+i>fl:i)=
1.3. Bii loin 3
R u t g Q n t 6 n g S = f + 2 ' + 3 ' + . . . + n' voi n e N ' . Hudng dan
Tim dugc da thuc f(x)= a.x'+b.x'+c.xdi f(x+l)-f(x)- x' Vx
Tacdfi;x+l>f(x)- x ' Vx
**
a.(x+l)=+b.(x+l)'+c.(x+l>(a.x'+b.x'+c.x)=x' Vx
•» 3a.x'+3ax+2b.x+a+b+c = x^ Vx _]_
~ 3
f(x)- a.x'+b.x'-h:.xd6 fl:x+l>f(x)- 3 x ' + 2 x + l Vx Tacd: f(x+l)-f(x)= 3x' + 2 x + l Vx
« • a(xH)=-lh(!tH)'-lc(xfl>(a^-lhx'-lcx)=3x'+2ii+l *
• » 3 a . x ' + 3 a x + 2 b J i + a t ^ t r t c = 3 x ' + 2 x + l Vx
3a = 3 3a+2b = 2 •»
a+b+c = l
b = - i =5-fl:x)=x'—x'+-!•!
2 2 2 c = — 1
2
3a = 1
« • 3a+2b=0«.
a+b+c = 0
b = -—=>fi'x)=-x^—\^ +-X 2 ' 3 2 6
"6 Tir d6 ta c6
S-H2)-fl;i>+fl:3>fi;2)+fl:4H3)+..+f(n+l)-f[n) = fi;n+l>f|;i)
=i.(„+.,4.(„+,)'+i(„+,)-(i.,..i.,+i,j
_ n(n+lX2n+l) 6 1.4. Bii toin 4
Rut gon tdng S = 6 + 1 7 + 34 + ... + ( 3 n ' + 2 n + l) vdi n e N * .
Htjdne dan
Khi gap bii tap niy ta cd thg tinh thdng qua kgt qua cua bii toin 2.
Nhung cung cd thg lim true tigp theo cich tigp can trgn.
Tim dugc da thiic:
Til dd ta cd
S= S:2>fl:i>tfl;3)-fl:2)+f(4K3)+-+l(n+l>f(n) - a:n+l)-([l)
=(n+l)=-i("+l)'+|("+'>-f''-l-''+|-')
n(2n'+5n+5) 2
Bing phuong phip trgn cho ta tdng quit bii toin Ihinh bii toin 1.5. Bii toin 5
Riit gon tdng S-{x+y+z)+ (4x+2y+z)+...+(xn'-t>iriz) vdi n e N*, X, y, z la cic sd thuc cho tnidc.
Khi cho bic cua sd hang tong quit Idn hon ta thu dugc cic bii toin mgi tdng quat hon niia. Tuy nhign bic cia da thtic fix) can tim ciing ting Ign. Vi du bii toin:
1.6. Bii toin 6
Rut ggn tdng S= 1 ' + 2 ' + 3 ' + . . . + n' vdi n e N ' . Hudng dan
Ta tim da thuc:
f(;x)= a.x''+b.x'+c.x'+dxdgf(x+l>f(x)= x' Vx Tacd:l(x+I)-f(x)= x' Vx
a(x+l)*+h(x+l)'+c(x+l)'-hi(x+l>(ax'+hx'-lM'+tk)=x' %
•» 4a.x'+«ax'+3b.x'+4a.x+3b.x+2c.x-fB+b+c+d=x' Vx
4a=l 6a+3b=0 4a+3b+2c=0
a+b+c+d=0 1 a=—
4
" T = i - f l : x ) = - ! - / - i x ' +-)?
, 4 2 4 c=— 4
d=0 Tir dd ta cd
S= fl:2>f(l)+fl:3)-fl;2)+fi:4>t[3)+.. +fl;n+l)-fl:n) = ffn+lKl)
= i ( n + l ) ' - i ( n + l ) ' + i ( n + l ) ' - f i . l ' - i . l = + i . l ' l 4 2 4 l4 2 4 J
(n+l)'.n'
D(yvaH^i 48
1.7. Bii toin 7
Rutgon tSng S-10 + 42 + 108 + ... + n(n+lX2n+3) vdi n e N ' .
Hudng din
Ta tim da thiic fl;x)= a.x*+b.x'+c.x'+dxdS ((x+I)- li;x)-x(x+lX2x+3) Vx
TacdfCx+l)-f(x)= 2x'+5x^+3x Vx
«a.(x+l)*+b.(x+l)'+c.(x+l)'+d{x+l>(a.x*+b.x'+c
«• 4a.x'+6a.x'+3b.x'+4a.x+3b.x+2c.x+a+b+c+d
=2x'+5x'+3x Vx
4 a ^ 6»+3b=5
«•
4»f3b+2c=3 atbtcM=0
a = -1 2
b= 3 „ , 1 . 2 , 1 J 2
=!.ftx)=-x +-X --X --X - 1 2 3 2 3 c=— 2
d=— - 2 3 Tir dd ta ed
S=8;2>«l)+fl:3>fi:2>tf(4)-fi;3>t..+f(n+l>f(n)-8:n+l)-f(l)
= | ( n + l ) * + | ( n + l ) ' - i ( n + l ) ^ | ( n + l )
_ i , . + 2,..i,^_2.,
[2 3 2 3 (n+l).[3(n+l)* +4(n+l)'-3(n+l)-4l
6 (n+l).n(n+2X3n+7)
6
Ta cd Ihl t6ng quit bii toin thinh 1.8. Bii toin 8
Rit ggn t6ng
S-(x+y+z+t)+ (8x+4y+2z+t)+...+(xn'+yn'+zn+t) vdi n 6 N*, x, y, z, t li eic s6 thuc cho tnrdc, 1.9. Bii toin 9
Ta l^i x6l vdi mdt bii toin quen thudc cua hgc sinh Idp 6 Riltggnt6ngS=l+3+3'+...+3'' vdi n € N ' Hudnp djin
Cich lim cua cic sich tham khio nhu sau TirS=:+3+3'+...+3"
suyra3S=3.(l+3+3'+...+3")= 3+3'+3'+...+3"*' Do dd ta cd:
3S-S- 3+3'+3'+...+3"*'-(l+3+3-+..+3") = 3"*'-l
m-
-1
Cieh giai trgn rit ngin ggn yi de higu doi vdi hgc sinh.
Tuy nhign ehiing ta vin sg tigp e§n bii toin vi tim hudng di nhu cic bii tip tren, vdi hi vgng cd thi md rdng dugc bii toin.
Tatlmf(x)dgn;x+l>f[x)=3'' Vx.
Phin tieh: Bk thu duoc dilu trgn ta chon f(x) cd chiia 3"
Khiadf{x+l)-f(x)-h(x+l).3"'-h(x).3''= 3" Vx.
•^ 3h(x+l)-h(x)= 1 Vx nen ta xgt h(x) bac 0 tire h(x)= a Tacd3h(x+l)-h(x)-l Vx
«• 3a-a=l-» 2a=l <=> a = - =!• fl;x)=-.3"
Tir dd ta cd
S- fl;i>f(0)+fi:2)-fl;i)+f(3>fi:2)+...+fi:n+l)-fi:n) = fln+l)- l j „ _ 3 - ' - l
2 ' 2 ^ 2
Vdi hudng di nhu trgn bii toin dugc giii quyit, nhung ngu s6 hang lAng quit ciia diy phiic tap hon thi cdn ip dung dugc khdng. Ta xgt bii tip sau
I.IO. BiitoinlO
Rlltggnt6ngS=1.3'+2.3'+3.3'+...+n.3" vdi n £ N ' . Hudng din
Tatlmfl;x)agfi;x+l)-fi:x)=x.3" Vx
Phin tieh: Di thu duge diiu trgn ta chgn f(x) cd ehira 3^
ey thi xgt fl;x)= h(x). 3"
Khi dd fi:x+l)-fi;x)=h(x+l). 3"" - h(x).3" - x.3" Vx.
*> 3h(x+l>h(x)- X Vx ngn ta xfl h(x) bie 1 nic h(x)- ax-H) Taed3h(x+l)-h(x)=x Vx
<* [a.(x+l)+b]3 - (a.x+b) = x Vx
«• 2a.x+3a+2b=x Vx
» 2a.x+3a+2b = X Vx
^ j 2a=l \ ''-
*''|3a+2b = 0** , >{{xH:
Til dd ta cd
S= f(2>fl:i)+fl:3)-f(2)+lt4)-fl:3)+.. .+fi:n+l)-f(n) = f(n+l>fi:i)
(2n-l)3"'+3
f4)-^'
Giio vien cin chii y cic em. Budc tim f(x) cd th^ chi ein lim d giiy nhip.
