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please help me with this:
 

What is the constant term in the expansion of \(\left(\sqrt{x}+\dfrac5x\right)^{9}\)?

 

I don't know quick tricks thanks for helping smiley

 Jan 27, 2022
edited by proyaop  Jan 27, 2022
edited by proyaop  Jan 27, 2022
 #1
avatar+117849 
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Hi proyaop,

 

 

The general term is

 

\(\displaystyle {\binom{9}{n}(\sqrt x)^{9-n}(\frac{5}{x})^n}\\~\\ =\displaystyle {\binom{9}{n}(x)^{\frac{9-n}{2}}(\frac{5}{x^n})}\\~\\ =\displaystyle {\binom{9}{n}*5(x)^{\frac{9-n-2n}{2}}}\\~\\ =\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\\)

You want the constant term so     

\(\frac{9-3x}{2}=0\\ 9-3x=0\\ 9=3x\\ x=3\)

 

So the contatn term is

\(=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\ =\displaystyle {\binom{9}{3}*5(x)^{\frac{9-3*3}{2}}}\\~\\ =\binom{9}{3}*5\\~\\ =84*5\\~\\ =420\)

 

You do need to check my working for careless mistakes of course. wink

 

LaTex:

\displaystyle {\binom{9}{n}(\sqrt x)^{9-n}(\frac{5}{x})^n}\\~\\
=\displaystyle {\binom{9}{n}(x)^{\frac{9-n}{2}}(\frac{5}{x^n})}\\~\\
=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-n-2n}{2}}}\\~\\
=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\

 Jan 27, 2022

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