Given:
[tex]\mleft(2x^2y^3-\frac{1}{4}y\mright)^9[/tex]
To find the 5th term:
Using the binomial expansion of
[tex](a+b)^n=\sum ^n_{i=0}\binom{n}{i}a^{\mleft(n-i\mright)}b^i[/tex]
Put n=9 and i=4 (Fifth term of the series)
We get,
[tex]\begin{gathered} \binom{9}{5}a^{(9-4)}b^4=\frac{9!}{(9-5)!5!}a^5b^4 \\ =\frac{9!}{4!5!}a^5b^4 \\ =\frac{6\times7\times8\times9}{4\times3\times2}a^5b^4 \\ =(7\times2\times9)a^5b^4 \\ =126a^5b^4\ldots\ldots\ldots(1) \end{gathered}[/tex]
Here, we have
[tex]\begin{gathered} a=2x^2y^3 \\ b=-\frac{1}{4}y \end{gathered}[/tex]
Substituting these values in (1) we get,
[tex]\begin{gathered} 126(2x^2y^3)^5(-\frac{1}{4}y)^4=126(2^5)x^{10}y^1^5(\frac{1}{4})^4y^4 \\ =\frac{126\times32}{256}x^{10}y^{15}y^4 \\ =15.75x^{10}y^{19} \end{gathered}[/tex]
Hence, the fifth term of the expansion is,
[tex]15.75x^{10}y^{19}[/tex]
