Difference between revisions of "1960 AHSME Problems/Problem 20"

(Problem)
(Solution)
Line 10: Line 10:
  
 
==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 <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 $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:

$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$

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. ! Missing delimiter (. inserted).).

We want the exponent of $x$ to be $7$, so \[2(8-n)-n=7\] \[16-3n=7\] \[n=3\]

If $n=3$, then the corresponding term is \[\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3\] \[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\] \[-14x^7\]

The answer is $\boxed{\textbf{(D)}}$.

See Also

1960 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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
All AHSME Problems and Solutions
Invalid username
Login to AoPS