Difference between revisions of "1950 AHSME Problems/Problem 16"
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== Problem== | == Problem== | ||
| − | The | + | 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> |
| − | + | ==Solution== | |
| − | = | + | Use properties of exponents to move the squares outside the brackets use difference of squares. |
| + | |||
| + | <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> | |
==See Also== | ==See Also== | ||
{{AHSME box|year=1950|num-b=15|num-a=17}} | {{AHSME box|year=1950|num-b=15|num-a=17}} | ||
Revision as of 21:18, 13 November 2011
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 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
