\begin{tabular}{ll}8 & (a) The expression of $$ (1+p x)^{n} $$ where $$ n0 $$ by the binomial theorem is \ $$ 1+14 x+21 p^{2} x^{2}+k x^{3}+\ldots $$, find the value of $$ p $$ and of $$ k $$. \ \hline (b) Evaluate the coefficient of $$ x^{5} $$ in the expansion of $$ \left(x^{2}-\frac{2}{x} ight)^{7} $$.\end{tabular}

Question

\begin{tabular}{ll}8 & (a) The expression of $$ (1+p x)^{n} $$ where $$ n>0 $$ by the binomial theorem is \ $$ 1+14 x+21 p^{2} x^{2}+k x^{3}+\ldots $$, find the value of $$ p $$ and of $$ k $$. \ \hline (b) Evaluate the coefficient of $$ x^{5} $$ in the expansion of $$ \left(x^{2}-\frac{2}{x}ight)^{7} $$.\end{tabular}

UpStudy AI · Accepted Answer

To solve the problem, we will break it down into two parts: (a) and (b).

### Part (a)

We need to express $$ (1 + px)^n $$ using the binomial theorem, which states:

$$
(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k
$$

In our case, we have $$ (1 + px)^n $$, which can be expanded as:

$$
(1 + px)^n = \sum_{k=0}^{n} \binom{n}{k} (px)^k = \sum_{k=0}^{n} \binom{n}{k} p^k x^k
$$

From the given expression $$ 1 + 14x + 21p^2x^2 + kx^3 + \ldots $$, we can identify the coefficients of $$ x^1 $$, $$ x^2 $$, and $$ x^3 $$.

1. **Coefficient of $$ x^1 $$**:
   $$
   \binom{n}{1} p = 14
   $$
   This simplifies to:
   $$
   n p = 14 \quad \text{(1)}
   $$

2. **Coefficient of $$ x^2 $$**:
   $$
   \binom{n}{2} p^2 = 21p^2
   $$
   This simplifies to:
   $$
   \frac{n(n-1)}{2} p^2 = 21p^2
   $$
   Assuming $$ p^2 \neq 0 $$, we can divide both sides by $$ p^2 $$:
   $$
   \frac{n(n-1)}{2} = 21 \quad \text{(2)}
   $$

3. **Coefficient of $$ x^3 $$**:
   $$
   \binom{n}{3} p^3 = k
   $$

Now, we will solve equations (1) and (2) to find $$ n $$ and $$ p $$.

From equation (2):
$$
n(n-1) = 42
$$
We can find integer solutions for $$ n $$:
- $$ n = 7 $$ gives $$ 7 \times 6 = 42 $$
- $$ n = -6 $$ is not valid since $$ n > 0 $$.

Now substituting $$ n = 7 $$ into equation (1):
$$
7p = 14 \implies p = 2
$$

Now we have $$ n = 7 $$ and $$ p = 2 $$.

Next, we can find $$ k $$:
$$
k = \binom{7}{3} p^3 = \binom{7}{3} (2^3) = 35 \times 8 = 280
$$

Thus, the values are:
- $$ p = 2 $$
- $$ k = 280 $$

### Part (b)

We need to evaluate the coefficient of $$ x^5 $$ in the expansion of $$ \left(x^2 - \frac{2}{x}\right)^7 $$.

Using the binomial theorem:
$$
\left(x^2 - \frac{2}{x}\right)^7 = \sum_{k=0}^{7} \binom{7}{k} (x^2)^{7-k} \left(-\frac{2}{x}\right)^k
$$

This simplifies to:
$$
= \sum_{k=0}^{7} \binom{7}{k} (-2)^k x^{2(7-k) - k} = \sum_{k=0}^{7} \binom{7}{k} (-2)^k x^{14 - 3k}
$$

We need the coefficient of $$ x^5 $$:
$$
14 - 3k = 5 \implies 3k = 9 \implies k = 3
$$

Now substituting $$ k = 3 $$ into the binomial expansion:
$$
\text{Coefficient} = \binom{7}{3} (-2)^3 = 35 \times (-8) = -280
$$

### Final Answers

(a) $$ p = 2 $$, $$ k = 280 $$

(b) The coefficient of $$ x^5 $$ is $$ -280 $$.
$$ p = 2 $$, $$ k = 280 $$, and the coefficient of $$ x^5 $$ is $$ -280 $$.

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