Question

find the 4th term expansion of (4x/5-5/2x)9

Answer 1

here is ur answer
hope it would help u

Answer 2

Answer:

$t_{4}= -\frac{84\times 2^{9}}{5^{3}}\times x^{3}$

Step-by-step explanation:

$\left(\frac{4x}{5}-\frac{5}{2x}\right)^{9}$

$t_{r+1}=^{n}C_{r}\:x^{n-r}\times a^{r}$

$t_{4}=t_{3+1}$

r=3, n = 9;

$t_{3+1}=^{9}C_{3}\times (\frac{4x}{5})^{9-3}\times (\frac{-5}{2x})^{3}$

$=\frac{9!}{(9-3)!3!}\times (\frac{4x}{3})^{6}\times\frac {(-5)^{3}}{(2x)^{3}}$

$=84\times \frac{4^{6}\times (-5)^{3}\times x^{6}}{5^{6}\times 2^{3} \times x^{3}}$

$=-\frac{84\times 2^{9}}{5^{3}}\times x^{3}$

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