Middle term of expansion of (x/3 - y)^8 ?

[Guest]

What is the middle term of the expansion (x/3-y)^8?

What would be the coefficient of x^7 in the expression (2x^3 + 1/5x^2)^9

thank you in advance

 

[galactus]

Please use parentheses. Your expressions are ambiguous.

\(\displaystyle (\frac{x}{3}-y)^{8}\;\ or\;\ (\frac{x}{3-y})^{8}?\)

 

[pka]

In (a+b)<SUP>8</SUP> has nine terms therefore the middle term is Ka<SUP>4</SUP> b<SUP>4</SUP>. We need to find the coefficient K=combin(8,4).

 

[soroban]

Re: Middle term of expansion

Hello, americo74!

I assume you're familiar with the Binomial Expansion . . .

What is the middle term of the expansion \(\displaystyle \L\left(\frac{x}{3}\,-\,y)^8\) ?

The middle term is: \(\displaystyle \L\,\begin{pmatrix}8\\4\end{pmatrix}\left(\frac{x}{3}\right)^4(-y)^4 \: = \;70\,\cdot\,\frac{x^4}{81}\,\cdot\,y^4 \:=\:\frac{70}{81}x^4y^4\)

What is the coefficient of \(\displaystyle x^7\) in the expression \(\displaystyle \left(2x^3\,+\,\frac{1}{5x^2{\right)^9\) ?

The only term that contains \(\displaystyle x^7\) is the fifth term.

\(\displaystyle \L\;\;\;\begin{pmatrix}9\\5\end{pmatrix}\left(2x^3\right)^5\left(\frac{1}{5x^2}\right)^4 \;= \;126\cdot 2^5\,\cdot\,x^{15}\,\cdot\,\frac{1}{5^4\cdot x^8} \;= \;\frac{126\cdot2^5}{5^4}\,x^7\)

Therefore, the coefficient is: \(\displaystyle \L\,\frac{4032}{625}\)

 

[Guest]

thanks to all who helped me