Question

Find the 5th term from the end in the expansion of (x^3/2-2/x^2)^6

Answer 1

The required ans is 60/x²

Answer 2

$Given \: binomial \:\Big(\frac{x^{3}}{2} - \frac{2}{x^{2}}\Big)^{6}$

$\blue { General \:term \: in \: expansion \:of}\\\blue { (x+a)^{n} \: is: }$

$\boxed { \pink { t_{r+1}= ^{n}C_{r} x^{n-r} a^{r} }}$

$t_{5} = t_{4+1}$

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

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

$= \frac{6!}{2! 4!} \times \Big( \frac{x^{6}}{2^{2}}\Big) \times \Big( \frac{2^{4}}{x^{8}}\Big)$

$= 15 \times \frac{1}{4} \times 16 \times \frac{1}{x^{2}}$

$= 15 \times 4 \times \frac{1}{x^{2}}$

$= \frac{60}{x^{2}}$

Therefore.,

$\red{5^{th} \:term \:in \:the \: expansion }\\\green {= \frac{60}{x^{2}} }$

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