Difference between revisions of "1950 AHSME Problems/Problem 16"
m |
m (small latex fix) |
||
| (4 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
| − | == Problem== | + | == Problem == |
The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is: | The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is: | ||
| Line 5: | Line 5: | ||
<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 | + | Using the binomial theorem, we can see that the number of terms is <math>\boxed{\mathrm{(B)}\ 5}</math>. |
| − | ==See Also== | + | == See Also == |
| − | {{AHSME box|year=1950|num-b=15|num-a=17}} | + | {{AHSME 50p box|year=1950|num-b=15|num-a=17}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:51, 11 October 2020
Problem
The number of terms in the expansion of ![$[(a+3b)^{2}(a-3b)^{2}]^{2}$](http://latex.artofproblemsolving.com/3/2/2/32205bbac02ec22aacf9ad06916b511e68ea2b2c.png)
when simplified is:
Solution
Use properties of exponents to move the squares outside the brackets use difference of squares.
![\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]](http://latex.artofproblemsolving.com/d/8/8/d88510de517abcc0d064612d89e7094e4470ec69.png)
Using the binomial theorem, we can see that the number of terms is 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| 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. 
