First slide

Binomial theorem for positive integral Index

Question

What is the middle term in the expansion of ${(\frac{x \sqrt{y}}{3} - \frac{3}{y \sqrt{x}})}^{12}$ ?

Moderate

A

$C (12, 7) x^{- 3} y^{3}$
B

$C (12, 6) x^{3} y^{- 3}$
C

$C (12, 7) x^{3} y^{- 3}$
D

$C (12, 6) x^{- 3} y^{3}$

Solution

Given expansion ${(\frac{x \sqrt{y}}{3} - \frac{3}{y \sqrt{x}})}^{12}$ is

We have general term in the expansion ${(x + a)}^{n}$

$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$

Here, index be the $n = 12$ middle term is ${(\frac{12}{2} + 1)}^{t h}$ i.e. $7^{t h}$ term.

$∴$ Required middle term $= T_{7}$

$T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$

$T_{7} = T_{6 + 1} =^{12} C_{6} {(\frac{x \sqrt{y}}{3})}^{6} {(- \frac{3}{y \sqrt{x}})}^{6}$ .

$T_{7} = T_{6 + 1} =^{12} C_{6} \frac{x^{6} y^{3}}{x^{3} y^{6}} =^{12} C_{6} x^{3} y^{- 3}$

$T_{7} = T_{6 + 1} =^{12} C_{6} x^{3} y^{- 3}$

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