If the coefficient of $x^2$ in the expansion of $(k+ \frac 1 3 x)^5$ is $30$. What is the value of the constant $k$?
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Use the binomial theorem. – David Jun 16 '14 at 23:59
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what is the easiest way to do the binomial theorem? – Roslyn Jun 17 '14 at 00:12
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See Andre Nicolas's answer. – David Jun 17 '14 at 00:27
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Hint: The term in $t^2$ in the binomial expansion of $(s+t)^5$ is $\binom{5}{2}s^3t^2$. So the term in $x^2$ in the expansion of $\left(k+\frac{x}{3}\right)^2$ is $\binom{5}{2}k^3 \frac{x^2}{9}$. It follows that the coefficient of $x^2$ is $\frac{10k^3}{9}$.
Remark: The full expansion of $(s+t)^5$ is $$\binom{5}{0}s^5t^0+\binom{5}{1}s^4t+\binom{5}{2}s^3t^2+\binom{5}{3}s^2t^3+\binom{5}{4}st^4+\binom{5}{5}s^0t^5,$$ or more simply $s^5+5s^4t+10s^3t^2+10s^2t^3+5st^4+t^5$.
In your course, the binomial coefficient $\binom{5}{2}$ may be called $C^5_2$. or ${}^5C_2$, or $C(5,2)$. And there are other notations!
André Nicolas
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once you do the binomial theorem, what do you do next?...like the process? – Roslyn Jun 17 '14 at 00:28
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To finish, set $\frac{10k^3}{9}=30$. Algebra gives $k^3=27$ so $k=3$. – André Nicolas Jun 17 '14 at 00:31