Difference between revisions of "1960 AHSME Problems/Problem 20"
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− | 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 <math>x</math> to be <math>7</math>, so | We want the exponent of <math>x</math> to be <math>7</math>, so | ||
Line 18: | Line 18: | ||
If <math>n=3</math>, 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> |
Revision as of 10:54, 20 December 2018
Problem
The coefficient of in the expansion of is:
Solution
By the Binomial Theorem, each term of the expansion is $\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}(\left\frac{-2}{x}\right)^n$ (Error compiling LaTeX. Unknown error_msg).
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 |