Difference between revisions of "1963 AHSME Problems/Problem 9"

Latest revision as of 15:40, 4 June 2018

In the expansion of $\left(a-\dfrac{1}{\sqrt{a}}\right)^7$ the coefficient of $a^{-\dfrac{1}{2}}$ is:

$\textbf{(A)}\ -7 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ -21 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 35$

By the Binomial Theorem, each term of the expansion is $\binom{7}{n}(a)^{7-n}(\frac{-1}{\sqrt{a}})^n$ .

We want the exponent of $a$ to be $-\frac{1}{2}$ , so $(7-n)-\frac{1}{2}n=-\frac{1}{2}$ $-\frac{3}{2}n = -\frac{15}{2}$ $n = 5$

If $n=5$ , then the corresponding term is $\binom{7}{5}(a)^{2}(\frac{-1}{\sqrt{a}})^5$ $-21a^{-\frac{1}{2}}$

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

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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

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 ==Solution==
-By the [[Binomial Theorem]], each term of the expansion is <math>\binom{7}{n}(a)^{7-n}(frac{-1}{\sqrt{a}})^n</math>.
+By the [[Binomial Theorem]], each term of the expansion is <math>\binom{7}{n}(a)^{7-n}(\frac{-1}{\sqrt{a}})^n</math>.
 We want the exponent of <math>a</math> to be <math>-\frac{1}{2}</math>, so
@@ Line 23: / Line 23: @@
 The answer is <math>\boxed{\textbf{(C)}}</math>.
 ==See Also==