Difference between revisions of "1950 AHSME Problems/Problem 16"

(Problem)
m (small latex fix)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
== Problem==
+
== Problem ==
  
The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{4} </math> when simplified is:
+
The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is:
  
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
  
==Solution==
+
== Solution ==
  
 
Use properties of exponents to move the squares outside the brackets use difference of squares.
 
Use properties of exponents to move the squares outside the brackets use difference of squares.
Line 11: Line 11:
 
<cmath>[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4</cmath>
 
<cmath>[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4</cmath>
  
Using the binomial theorem, we can see that the number of terms is <math>\boxed{\mathrm{(B)}\ 5.}</math>
+
Using the binomial theorem, we can see that the number of terms is <math>\boxed{\mathrm{(B)}\ 5}</math>.
  
==See Also==
+
== See Also ==
  
 
{{AHSME 50p box|year=1950|num-b=15|num-a=17}}
 
{{AHSME 50p box|year=1950|num-b=15|num-a=17}}

Latest revision as of 23:51, 11 October 2020

Problem

The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

Use properties of exponents to move the squares outside the brackets use difference of squares.

\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]

Using the binomial theorem, we can see that the number of terms is $\boxed{\mathrm{(B)}\ 5}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png