Question 7 The term $$ 10 x^{3} y^{2} $$ is a term in which binomial expansion? $$ \frac{(x+y)^{6}}{(x+y)^{8}} $$ $$ \frac{(x+y)^{5}}{(5 x+2 y)^{6}} $$

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To determine which binomial expansion the term $$ 10x^3y^2 $$ belongs to, we need to compare the given term with the general term of the binomial expansion of each expression.

The general term of the binomial expansion of $$(x+y)^n$$ is given by:
$$ T_{r+1} = \binom{n}{r} x^{n-r} y^r $$

Let's analyze each given expression:

1. $$\frac{(x+y)^6}{(x+y)^8}$$
This expression simplifies to $$(x+y)^{-2}$$. The general term of $$(x+y)^{-2}$$ is:
$$ T_{r+1} = \binom{-2}{r} x^{-2-r} y^r $$
Since $$\binom{-2}{r}$$ is not defined for all $$r$$, this expression does not have a standard binomial expansion.

2. $$\frac{(x+y)^5}{(5x+2y)^6}$$
This expression does not simplify to a binomial expansion in the form $$(x+y)^n$$. It involves a different binomial $$(5x+2y)$$, so it is not a standard binomial expansion.

Now, let's find the binomial expansion that includes the term $$10x^3y^2$$.

The term $$10x^3y^2$$ can be rewritten as:
$$ 10x^3y^2 = 10 \cdot x^3 \cdot y^2 $$

We need to find the binomial expansion where the general term matches this form. The general term of $$(x+y)^n$$ is:
$$ T_{r+1} = \binom{n}{r} x^{n-r} y^r $$

For the term $$10x^3y^2$$ to be present, we need:
$$ n-r = 3 $$
$$ r = 2 $$

This means that the term $$10x^3y^2$$ is the fourth term (since $$r+1 = 4$$) in the expansion of $$(x+y)^n$$, where $$n-r = 3$$ and $$r = 2$$. Solving for $$n$$, we get:
$$ n - 2 = 3 $$
$$ n = 5 $$

Therefore, the term $$10x^3y^2$$ is a term in the binomial expansion of $$(x+y)^5$$. The correct answer is:
$$ \frac{(x+y)^5}{(x+y)^5} $$
This expression simplifies to 1, but it is the correct binomial expansion that includes the term $$10x^3y^2$$.
The term $$10x^3y^2$$ is a term in the binomial expansion of $$(x+y)^5$$.

Question 7 The term \( 10 x^{3} y^{2} \) is a term in which binomial expansion? \( \frac{(x+y)^{6}}{(x+y)^{8}} \) \( \frac{(x+y)^{5}}{(5 x+2 y)^{6}} \)

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