First slide

Binomial theorem for positive integral Index

Question

The 8th term of ${(3 x + \frac{2}{3 x^{2}})}^{12}$ when expanded in ascending power of x, is

Moderate

A

$\frac{228096}{x^{3}}$
B

$\frac{228096}{x^{9}}$
C

$\frac{328179}{x^{9}}$
D

none of these

Solution

When ${(3 x + \frac{2}{3 x^{2}})}^{12}$ is expanded, the power of x goes on decreasing as the terms proceed.
Hence, it is expanded in descending powers of x. So ${(\frac{2}{3 x^{2}} + 3 x)}^{12}$ ,when expanded, will be in ascending powers of x.
Now,
$t_{8} in {(\frac{2}{3 x^{2}} + 3 x)}^{12} =^{12} C_{7} {(\frac{2}{3 x^{2}})}^{12 - 7}$
$\begin{aligned} = \frac{12!}{7! 5!} \cdot {(\frac{2}{3 x^{2}})}^{5} \cdot (3 x)^{7} \\ = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2} \cdot \frac{2^{5} \cdot 3^{2}}{x^{3}} \\ = \frac{228096}{x^{3}} \end{aligned}$

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