1 what is the com efficient of x3in the expansion of ( 1+x) 5 2 in the expansion of (2x- 3)4, what is the coefficient of x2 3 The term independent of (x) in the expansion of (1-2x) is 4 In the expansion of (a-b)n the sum of coefficient is? 5 The middle term in the expansion of ( 1-2x) -4

The general term in the binomial expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
For $(1+x)^5$, we have $a=1$, $b=x$, and $n=5$.
We want the term with $x^3$, so we set $k=3$.

$$T_{3+1} = \binom{5}{3} (1)^{5-3} (x)^3$$ $$T_4 = \binom{5}{3} (1)^2 (x)^3$$

Calculate $\binom{5}{3}$:

$$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$$

So the term is $10 \cdot 1 \cdot x^3 = 10x^3$.
The coefficient of $x^3$ is 10.

The coefficient of $x^3$ is $\boxed{\text{10}}$.

The general term in the binomial expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
For $(2x-3)^4$, we have $a=2x$, $b=-3$, and $n=4$.

$$T_{k+1} = \binom{4}{k} (2x)^{4-k} (-3)^k$$

We want the term with $x^2$, so the power of $x$ must be 2.
From $(2x)^{4-k}$, the power of $x$ is $4-k$.
Set $4-k=2$:

$$4-k=2 \implies k=2$$

Now substitute $k=2$ into the general term:

$$T_{2+1} = \binom{4}{2} (2x)^{4-2} (-3)^2$$ $$T_3 = \binom{4}{2} (2x)^2 (-3)^2$$

Calculate $\binom{4}{2}$:

$$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$$

Substitute the values:

$$T_3 = 6 \cdot (4x^2) \cdot (9)$$ $$T_3 = 6 \cdot 4 \cdot 9 \cdot x^2$$ $$T_3 = 216x^2$$

The coefficient of $x^2$ is 216.

The coefficient of $x^2$ is $\boxed{\text{216}}$.

The expression given is $(1-2x)$.
The term independent of $x$ is the constant term in this expression.
The constant term is 1.

The term independent of $x$ is $\boxed{\text{1}}$.

To find the sum of coefficients in any polynomial expansion, substitute 1 for each variable.
In the expansion of $(a-b)^n$, we substitute $a=1$ and $b=1$.
Sum of coefficients $= (1-1)^n = 0^n$.
If $n > 0$, then $0^n = 0$.
If $n=0$, then $(a-b)^0 = 1$, and the sum of coefficients is 1.
Assuming $n$ is a positive integer, which is typical for binomial expansion questions of this type.

The sum of coefficients is $\boxed{\text{0}}$.

The expression $(1-2x)^{-4}$ represents an infinite series expansion, as the exponent is a negative integer.
For example, using the generalized binomial theorem $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$
Here, $y = -2x$ and $n = -4$.

$$(1-2x)^{-4} = 1 + (-4)(-2x) + \frac{(-4)(-4-1)}{2!}(-2x)^2 + \dots$$ $$= 1 + 8x + \frac{(-4)(-5)}{2}(4x^2) + \dots$$ $$= 1 + 8x + 10(4x^2) + \dots$$ $$= 1 + 8x + 40x^2 + \dots$$

An infinite series does not have a "middle term" because there is no finite number of terms to determine a central position.

The expansion of $(1-2x)^{-4}$ is an infinite series and therefore has $\boxed{\text{no middle term}}$.