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203 
THE BINOMIAL THEOREM. 



By G. B. M. ZKHE, A. M., Ph. D., Texarkana, Tezas. 



I use the following rule for expanding all binomials, whether the expon- 
ent is integral or fractional, positive or negative. 

The number of terms of a binomial expansion is one more than the expon- 
ent when the exponent is a positive integer, otherwise the number of terms is in- 
finite. For the first term of the expansion, raise the first term of the binomial 
to the required power. For any other term of the expansion, multiply the pre- 
ceding term by the second term of the binomial, and this product by the expon- 
ent of the power diminished by two less than the number of terms from the be- 
ginning, divide this product by the product of the first term of the biuomial into 
one less than the number of terms from the beginning, always observing 
the proper algebraic signs of the binomial terms. 

i ,-u-^m i » , m(ax) m by , m(jn—V){ax) m (by) i , 

(rw+ byy= (ax)' + ^ JL + ^T&e) 8 

w(m-l)(m-2) (m-r+2)(a30'"(fey)'-' 

+ + 1.2.3 (r-l)(axy-t + U; * 

(.4) gives the expansion without reducing the terms. 
(1). To expand (3x±4y) s . 

1st term =(3x) 5 =243a; s ; 2nd term = 24Sx " X g ±4y)x5 =±162(k«2/ ; 

„■,. ±1620a;* V x(±4y)x4 ..,„,. 
3rd term = 1—^ — ^ — =4d20x 3 y t ; 

.... 4320zyx(±4v)x3 ,«--„„•„, 
4th term = ^— ~ ^ =±5760a; 2 v 3 ; 

,,. . ±5760x*y 3 x(±4y)x2 noin , 
5th term = ^— -i y ' =3840a;y« ; 

„ . , . 3840z V 4 x (±4y) x 1 , , nn . . 
6th term == ^~c~^ — ===t: 10243/ 5 . 

.-. (3a;±4i/) ,! =;243a; e ± 162Cte*y+4320a:V ± 5760a;*y 3 +3840xj/ i ±1024i/ 6 . 



204 

(2) . To expand (a * -f 26) ' . 



1st term =(a*y=a**; 2nd term = a ' 4 -f ) - 7 ==14a i *b ; 



_,, 14a' ! 6.2b.6 .. liIt ... . 84a 1 »b 8 .26.5 oan ... 

3rd term = s — 5 =84a' "6 s ; 4th term = — — s — ; ^280a*6 3 ; 

2. a* 3. a s 

_ iU , 280a 8 b 9 .2b.4 ... ... „ iU . 560a 6 6 4 .2b.3 „-„.., 

5th term = 3 — 5 =5600" 6* ; 6th term = = — 5 =672a 4 b 5 ; 

4. a* 5. a s 

„... 672a 4 b«.26.2 ... ... ... . 448a 8 6«.26.1 ,„_,, 

7th term = 5 — ; =448o ! 6 8 ; 8th term = =— -„ =128b 1 . 

.-. (a 8 +2b)' l =a 14 + l4a l *&+84a 1,, 6 8 +280a 8 & s 

+ 560a'6 4 +672a 4 & 5 + 448^" + 128b 1 . 

(3). To expand (2 + x)-» . 

1st term =(2)-* =4 ; 2nd term --={ x zx( ~ 3) = - -J|_; 

u lb 

„,, 3x xx(-4) 3a; 8 ... , 3a; 8 xx(-5) 5x' 

3rd term =-- jg-X -^- = ^5 4th term =_x -g^-W- — ; 

... . '5a; 3 xx (-6) 15x 4 

0thterm== -32- X --4X--l28- 

. . C^+x) _ a 16 + 16 32 + 128 



2r 
(4). To expand (1 + =£-)' . 

3 ° J 
1st term =(1)5 =1 ; 2nd term = =x ; 

3rd term = — ^ — =ix« ; 4th term = — =—hx>> ■ 

5th term = ^ =rbK*. 



205 



.-. (1 + ^)1=1 + ^ + ^-^3+^^. 



(5). To expand (8 + 12a)? . 



1st term =(8)' =4 ; 2nd term = ' — =4a ; 



3 



„ , . 4a.l2a.(-i) . , t . , -aM2a.(-J) 2a 
3rd term = ^ '-==— a 8 ; 4th term =■ ^-5 — =— 5- , 

Z.O O.O o 

Q„3 

5th term = ^ - -^. 

.-. (8 + 12a)»=4+4a-a i! + §a 3 - l y» 4 + 

(6). To expand (4a- 8x)-» . 

1st term =(4a ) - i =^; 2nd term^. ^g^> = •£; 

3rd term = « - ^Mzij = ^ 4t h term *?. ^=j) = *f . 
2aS 2.4a 4a$ ' 4a? 3.4a 4a? ' 

5x s (-8x)(-J) 35** 
5th term = -r-r. .'; l ' = -j^. 

4a* 4.4a loa» 

/.(4a-8a)-l==^- r +-g- r +-f- r +- i - r +- 



2ai 7 2a* 4a? T 4a? ^ 16a! 



1 / S 3x» 5x» 35a;* 
2a* V + a 2a 8+ 4a 3+ 8a* + 



The r"> term in \&\ j ^™" 1 )^- 2 ) (m-r+2)(axH%)^ 

lhe f term m U) is 1 2 - ir _ 1){ax y-i 



(7) . Find the 4th term of (-^-+9&) 10 . 



10x9x8x(-|-) ,0 (9&) 3 

m=10, r=4. .-. 4th term = =40a 1 t 3 . 

1.2>3.(-|-) 3 



206 



(8). Find the 28th term of (5a; + Si/)' ° . 
m=30, r=28. 

• 2Sth term- 30 ' 29 - 28 6.5.4.(5^ »(8y)" _ IjO m * m „ 

. .^Sthterm- 123 2 6.27.(5x)*< ~ I 27 JJ (M > (0y) ' 

(9). Find the 8th term of (1 + 2a-)-t . 
m— — I, r=8. 

• • 8th term ~ L273.4.5.6.7.C1) 1 ~ 16 " 

(10). Find the 10th term of (l + 3a 8 )V. 
m=V. r=10. 

■ 10th tnnn- V-V-V-M-K-D^-tJ^-IXi^aa')' , 1040a' 8 

• • 1UUl term " 1.2.3.4.5.6.7.8.9.(1)' ~ &T~' 

(11). Find the 5th term of (3a-26)-' . 
m=— 1, r=5. 

,.5th terra= (zi)(-2)(-3)(-4)(3a)-(2b)^ 166* 



1.2.3.4.(3a)« 243a 6 ' 

These are enough examples to illustrate both the rule and the general 
term. 

I have used this method with my classes for several years and find 
it easier and better than any other method I have ever used. I have never seen 
this method in this form. If any of the readers of the Monthly have ever seen 
it, I would be pleased to know where to find it.