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Find the constant term in the expansion of $$\left(10x^3-\frac{1}{2x^2}\right)^{5}$$

 Sep 11, 2018
 #1
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expand  (10 x^3 - 1/(2 x^2))^5=100,000 x^15 - 25,000 x^10 - 1/(32 x^10) + 2,500 x^5 + 25/(8 x^5) - 125 - the constant term.

 Sep 11, 2018
 #2
avatar+20842 
+7

Find the constant term in the expansion of $$\left(10x^3-\frac{1}{2x^2}\right)^{5}$$

 

\(\begin{array}{|rcll|} \hline && \left(10x^3-\dfrac{1}{2x^2}\right)^{5} \\\\ &=& \left(\dfrac{10x^3\cdot 2x^2-1}{2x^2}\right)^{5} \\\\ &=& \left(\dfrac{20x^5-1}{2x^2}\right)^{5} \\\\ &=&\dfrac{1}{\left(2x^2 \right)^5} \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 2^5x^{10} } \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 32x^{10} } \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 32x^{10} } \left[ \dbinom{5}{0}\left(20x^5 \right)^5 -\dbinom{5}{1}\left(20x^5 \right)^4 +\dbinom{5}{2}\left(20x^5 \right)^3 \\ -\dbinom{5}{3}\left(20x^5 \right)^2 +\dbinom{5}{4}\left(20x^5 \right) +\dbinom{5}{5}(-1)^5 \right] \\\\ &=&\dfrac{1}{ 32x^{10} } \left[ \dbinom{5}{0} 20^5x^{25} -\dbinom{5}{1} 20^4x^{20} +\dbinom{5}{2} 20^3x^{15} \\ { \color{red}-\dbinom{5}{3} 20^2x^{10} } +\dbinom{5}{4} 20x^5 -\dbinom{5}{5} \right] \\ \hline \end{array}\)

 

Constant term:

\(\begin{array}{|rcll|} \hline && \dfrac{1}{ 32x^{10} }\left[ { \color{red}-\dbinom{5}{3} 20^2x^{10} } \right] \\\\ &=& -\dfrac{1}{ 32x^{10} } \dbinom{5}{3} 20^2x^{10} \\\\ &=& -\dfrac{20^2}{ 32 } \dbinom{5}{3} \quad & | \quad \dbinom{5}{3} = \dbinom{5}{5-3}=\dbinom{5}{2}=\dfrac{5}{2}\cdot \dfrac{4}{1} = 10 \\\\ &=& -\dfrac{20^2\cdot 10}{ 32 } \\\\ &=& -\dfrac{4000}{ 32 } \\\\ &\mathbf{=}& \mathbf{-125} \\ \hline \end{array}\)

 

laugh

 Sep 12, 2018

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