Question

Find the middle term in the expansion (3x/7-2y)^10

Answer 1

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Answer 2

The middle term in expansion of $\left( {\frac{3x}{7} - 2y} \right)^{10}$ is

$\bf \red{ \: T_{6}= - 116.59 \: {x}^{5} {y}^{5} }$

Step-by-step explanation:

Given:

• $\left( {\frac{3x}{7} - 2y} \right)^{10}$

To find:

• Find the middle term of the expansion.

Solution:

Concept/Formula to be used:

1. General term in binomial expansion:In the binomial expansion of $( {a + b)}^{n}$, general term is given by $\bf T_{r+1}=^ nC_r a^{n-r}b^r$
2. Middle term : If n is even, then $\left(\frac{n}{2} + 1 \right)^{th}$term is middle term.
3. $^nC_r = \frac{n!}{r!(n - r)!}$

Step 1:

Find the middle term.

Here, in $\left( {\frac{3x}{7} - 2y} \right)^{10}$

n=10, even

So, term $\left(\frac{10}{2} + 1 \right)^{th}$

or

$\bf {6}^{th}$term is middle term in the expansion.

Step 2:

Find the 6th term.

From the formula of general term, we can find 6th term.

Middle term of the expansion is

put r=5

$T_{6}=^ {10}C_5 \left( \frac{3x}{7} \right)^{5}( - 2y)^5$

Simplify

$T_{6}= - \frac{10!}{5!5!} \left( \frac{243{x}^{5} }{16807} \right) \times 32 {y}^{5}$

or

$T_{6}= - \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \left( \frac{243{x}^{5} }{16807} \right) \times 32 {y}^{5}$

or

$T_{6}= - 252 \left( \frac{243{x}^{5} }{16807} \right) \times 32 {y}^{5}$

or

$T_{6}= - \frac{1959552}{16807} {x}^{5} {y}^{5}$

or

$\bf \: T_{6}= - 116.59 \: {x}^{5} {y}^{5}$

Thus,

The middle term in expansion of $\left( {\frac{3x}{7} - 2y} \right)^{10}$ is

$\bf \: T_{6}= - 116.59 \: {x}^{5} {y}^{5}$

Learn more about binomial expansion:

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