Difference between revisions of "1960 AHSME Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | The coefficient of <math>x^7</math> in the expansion of <math>(\frac{x^2}{2}-\frac{2}{x})^8</math> is: | + | The coefficient of <math>x^7</math> in the expansion of <math>\left(\frac{x^2}{2}-\frac{2}{x}\right)^8</math> is: |
<math>\textbf{(A)}\ 56\qquad | <math>\textbf{(A)}\ 56\qquad | ||
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==Solution== | ==Solution== | ||
− | By the Binomial Theorem, each term of the expansion is <math>\binom{8}{n}(\frac{x^2}{2})^{8-n}(\frac{2}{x})^n</math>. | + | By the [[Binomial Theorem]], each term of the expansion is <math>\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}\left(\frac{-2}{x}\right)^n</math>. |
− | We want the exponent of | + | We want the exponent of <math>x</math> to be <math>7</math>, so |
<cmath>2(8-n)-n=7</cmath> | <cmath>2(8-n)-n=7</cmath> | ||
<cmath>16-3n=7</cmath> | <cmath>16-3n=7</cmath> | ||
<cmath>n=3</cmath> | <cmath>n=3</cmath> | ||
− | If n=3, then the corresponding term is | + | If <math>n=3</math>, then the corresponding term is |
− | <cmath>\binom{8}{3}(\frac{x^2}{2})^{5}(\frac{-2}{x})^3</cmath> | + | <cmath>\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3</cmath> |
<cmath>56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}</cmath> | <cmath>56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}</cmath> | ||
<cmath>-14x^7</cmath> | <cmath>-14x^7</cmath> | ||
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==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1960|num-b=19|num-a=21}} | {{AHSME 40p box|year=1960|num-b=19|num-a=21}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 11:55, 20 December 2018
Problem
The coefficient of in the expansion of is:
Solution
By the Binomial Theorem, each term of the expansion is .
We want the exponent of to be , so
If , then the corresponding term is
The answer is .
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AHSME Problems and Solutions |