Question

The third term from the end in the expansion of ${(\frac{3x}{5}-\frac{5}{2x})}^8$ is

• A. $\frac{35451}{15x^4}$
• B. $\frac{45455}{16x^4}$
• C. $\frac{39372}{15x^4}$
• D. $\frac{39375}{16x^4}$

Answer 1

Answer: (D)

$\frac{39375}{16x^4}$

$Thirdtermfromtheendwillbe(8-3+2)thtermfrombeginning{T}_7{=}^8{C}_6\left(\frac{3x}{5}{)}^2\right(\frac{-5}{2x}{)}^6=\left(\frac{8\times 7}{2}\right)\times \left(\frac{9}{25}\right)\times x^2\times \left(\frac{15625}{64\times x^6}\right)=\left(\frac{39375}{16x^4}\right)$