The 5th term in the expansion is [tex]15.75x^{10}y^{19}[/tex]
Explanation:
The expression is [tex](2x^2y^3-\frac{1}{4})^9[/tex]
We need to determine the 5th term of the expansion.
The expression can be expanded using the binomial theorem.
Hence, we have,
[tex](2x^2y^3-\frac{1}{4})^9=9C0 (2x^{2} y^3)^{9-0}(-\frac{1}{4}y)^0+ 9C1 (2x^{2} y^3)^{9-1}(-\frac{1}{4}y)^1+ 9C2 (2x^{2} y^3)^{9-2}(-\frac{1}{4}y)^2+ 9C3 (2x^{2} y^3)^{9-3}(-\frac{1}{4}y)^3+ 9C4 (2x^{2} y^3)^{9-4}(-\frac{1}{4}y)^4+ 9C5 (2x^{2} y^3)^{9-5}(-\frac{1}{4}y)^5+ 9C6 (2x^{2} y^3)^{9-6}(-\frac{1}{4}y)^6+ 9C7 (2x^{2} y^3)^{9-7}(-\frac{1}{4}y)^7+ 9C8 (2x^{2} y^3)^{9-8}(-\frac{1}{4}y)^8+ 9C9 (2x^{2} y^3)^{9-9}(-\frac{1}{4}y)^9[/tex]The 5th term in the binomial expansion is [tex]9C4 (2x^{2} y^3)^{9-4}(-\frac{1}{4}y)^4[/tex]
Now, we shall simplify the expansion.
Thus, we have,
[tex]126 (2x^{2} y^3)^{5}(-\frac{1}{4}y)^4[/tex]
Multiplying the power, we get,
[tex]126 (32x^{10} y^{15})(\frac{1}{256}y^4)[/tex]
Simplifying the powers, we have,
[tex]\frac{4032x^{10}y^{19}}{256}[/tex]
Dividing, we get,
[tex]15.75x^{10}y^{19}[/tex]
Thus, the 5th term in the expansion is [tex]15.75x^{10}y^{19}[/tex]
