Difference between revisions of "1950 AHSME Problems/Problem 16"
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− | == 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: | ||
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<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. | ||
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<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 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 when simplified is:
Solution
Use properties of exponents to move the squares outside the brackets use difference of squares.
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 | |
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All AHSME Problems and Solutions |
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