Let p be a real constant and n be a positive integer.
(a) Expand    6 1 1n
px x    in ascending powers of x as far as the term in 3
x .
(b) If the coefficients of x and 2
x are 17 and 50, find the coefficient of 3
x .

Let's break down this problem step by step:

### Part (a): Expand \((1 - px)^{-n}(1 + x)^6\) in ascending powers of \(x\) as far as the term in \(x^3\).

We will use the binomial expansion for both terms separately and then multiply them together.

#### Step 1: Expand \((1 - px)^{-n}\)

Using the binomial expansion for \((1 - px)^{-n}\), we get:

\[
(1 - px)^{-n} = 1 + (-n)(-px) + \frac{-n(-n-1)}{2!}(px)^2 + \frac{-n(-n-1)(-n-2)}{3!}(px)^3 + \cdots
\]

Simplifying the terms:

\[
(1 - px)^{-n} = 1 + npx + \frac{n(n+1)}{2}p^2x^2 + \frac{n(n+1)(n+2)}{6}p^3x^3 + \cdots
\]

#### Step 2: Expand \((1 + x)^6\)

Using the binomial expansion for \((1 + x)^6\), we get:

\[
(1 + x)^6 = 1 + 6x + \frac{6(5)}{2}x^2 + \frac{6(5)(4)}{6}x^3 + \cdots
\]

Simplifying the terms:

\[
(1 + x)^6 = 1 + 6x + 15x^2 + 20x^3 + \cdots
\]

#### Step 3: Multiply the expansions

Now, multiply the two expansions term by term, up to the \(x^3\) term.

\[
\text{Result} = (1 + npx + \frac{n(n+1)}{2}p^2x^2 + \frac{n(n+1)(n+2)}{6}p^3x^3) \times (1 + 6x + 15x^2 + 20x^3)
\]

Let's expand this product term by term:

1. **Constant term:**

\[
1 \times 1 = 1
\]

2. **Coefficient of \(x\):**

\[
1 \times 6x + npx \times 1 = 6x + npx
\]

So, the coefficient of \(x\) is \(6 + np\).

3. **Coefficient of \(x^2\):**

\[
1 \times 15x^2 + npx \times 6x + \frac{n(n+1)}{2}p^2x^2 \times 1
\]
\[
= 15x^2 + 6npx^2 + \frac{n(n+1)}{2}p^2x^2
\]

So, the coefficient of \(x^2\) is \(15 + 6np + \frac{n(n+1)}{2}p^2\).

4. **Coefficient of \(x^3\):**

\[
1 \times 20x^3 + npx \times 15x^2 + \frac{n(n+1)}{2}p^2x^2 \times 6x + \frac{n(n+1)(n+2)}{6}p^3x^3 \times 1
\]
\[
= 20x^3 + 15npx^3 + 6\frac{n(n+1)}{2}p^2x^3 + \frac{n(n+1)(n+2)}{6}p^3x^3
\]

Simplifying, the coefficient of \(x^3\) is:

\[
20 + 15np + 3n(n+1)p^2 + \frac{n(n+1)(n+2)}{6}p^3
\]

### Part (b): Solve for \(p\) and find the coefficient of \(x^3\)

We are given that the coefficient of \(x\) is \(-17\) and the coefficient of \(x^2\) is \(50\).

1. **For the coefficient of \(x\):**

\[
6 + np = -17
\]

Solving for \(np\):

\[
np = -17 - 6 = -23
\]

So, \(np = -23\).

2. **For the coefficient of \(x^2\):**

\[
15 + 6np + \frac{n(n+1)}{2}p^2 = 50
\]

Substitute \(np = -23\):

\[
15 + 6(-23) + \frac{n(n+1)}{2}p^2 = 50
\]
\[
15 - 138 + \frac{n(n+1)}{2}p^2 = 50
\]
\[
-123 + \frac{n(n+1)}{2}p^2 = 50
\]
\[
\frac{n(n+1)}{2}p^2 = 50 + 123 = 173
\]
\[
n(n+1)p^2 = 346
\]

Now we have two equations:
1. \(np = -23\)
2. \(n(n+1)p^2 = 346\)

From \(np = -23\), solve for \(p\):

\[
p = \frac{-23}{n}
\]

Substitute this into \(n(n+1)p^2 = 346\):

\[
n(n+1)\left(\frac{-23}{n}\right)^2 = 346
\]
\[
n(n+1)\frac{529}{n^2} = 346
\]
\[
(n+1)\frac{529}{n} = 346
\]
\[
529(n+1) = 346n
\]
\[
529n + 529 = 346n
\]
\[
529 = -183n
\]
\[
n = \frac{529}{183} \approx 2.89
\]

Since \(n\) must be a positive integer, this result suggests there may be additional constraints that need reevaluation based on rounding errors or misinterpretation of terms.

Let me correct details, but does this satisfy?



Let p be a real constant and n be a positive integer. (a) Expand (1 - px)^-n(1 + x)^6 in ascending powers of x as far as the term in x^3. (b) If the coefficients of x and x^2 are -17 and 50, find the coefficient of x^3.

Explore the step-by-step solution to expanding (1 - px)^-n(1 + x)^6 in powers of x up to the term in x^3. Learn how the binomial theorem is used to solve for the coefficients of x, x^2, and x^3, given specific conditions such as coefficients -17 and 50.

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