Question

find the 7th term in the expansion of (1-x^2/3)^-4​

Answer 1

Step-by-step explanation:

The given term is (3x^{2}-\frac{1}{3} )^1^0,

Binomial expansion of (from above given term x=3x² and b==31 and n=10;

the general term of the binomial expansion is give by Ta+1="C₂xn-aya.. (1)

therefore, the 7th term can be found bu replaying a=6 in equation (1)

therefore, the 7th term is T,= ¹°С6x10-6y6

therefore, ¹°C6= 10! (10-6)!(6)!

¹⁰C6 = 24

Thus the 7th term of the given binomial is 24x4y6

Answer 2

Answer:

Concept:

Finding the term from the given form by using the Binomial theorem.

Step-by-step explanation:

Given:

$1-\frac{x^{2}}{3}\right)^{-4}$

Find:

$7^{\text {th }} \text { term of }\left(1-\frac{x^{2}}{3}\right)^{-4}$

Solution:

Now, we need to take

$T_{r}+1\text { in }\left(1-{ x }\right) \ ^{\text{-n}}$

that is    $\frac{n(n+1)(n+2) \cdots(n+r-1)}{1 \cdot 2 \cdot 3 \cdots-r} .$$x^r$

because the given equation is in the form of  $(1-{ x }\right)) \ ^{\text{-n}}$

Substitute the values according to the formula and given equation,

$r=6, \ n=4, x=\frac{x^{2}}{3}$

$7^{\text {th }} \text { term }=\frac{4(4+1)(4+2)(4+3)(4+4)(4+5)}{1 \cdot 2 \cdots 6}\left(-\frac{x^{2}}{3}\right)^{6}$

$=\frac{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} \times\left(\frac{x^{2}}{3}\right)^{6}$

$=\frac{28}{243} x^{12}$

So the answer is    $\frac{28}{243} x^{12}$

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