Write down, and simplify, the first 4 terms in the expansion of in ascending powers
of .
Replacing by , determine the coefficient of in the expansion of

Ask by Moran Paul. in Singapore

Mar 23,2025

Question

Write down, and simplify, the first 4 terms in the expansion of in ascending powers
of .
Replacing by , determine the coefficient of in the expansion of

Ask by Moran Paul. in Singapore

Mar 23,2025

UpStudy AI · Accepted Answer

We start with the expansion of $$(1+x)^9$$. By the Binomial Theorem, we have

$$
(1+x)^9 = \sum_{k=0}^{9} \binom{9}{k} x^k.
$$

The first 4 terms (i.e. terms with $$x^0$$, $$x^1$$, $$x^2$$, and $$x^3$$) are:

$$
\begin{aligned}
\text{Term when } k=0: &\quad \binom{9}{0} x^0 = 1, \
\text{Term when } k=1: &\quad \binom{9}{1} x^1 = 9x, \
\text{Term when } k=2: &\quad \binom{9}{2} x^2 = 36x^2, \quad (\text{since } \binom{9}{2} = 36), \
\text{Term when } k=3: &\quad \binom{9}{3} x^3 = 84x^3, \quad (\text{since } \binom{9}{3} = 84).
\end{aligned}
$$

Thus, the expansion (up to the $$x^3$$ term) is

$$
(1+x)^9 = 1 + 9x + 36x^2 + 84x^3 + \cdots.
$$

Next, we replace $$x$$ with $$z-z^2$$ so that

$$
(1+z-z^2)^9 = \sum_{k=0}^{9} \binom{9}{k} (z-z^2)^k.
$$

We are interested in finding the coefficient of $$z^3$$ in this expansion.

For a given term with index $$k$$, we expand $$(z-z^2)^k$$ by choosing $$i$$ factors of $$z$$ and $$k-i$$ factors of $$-z^2$$. This gives

$$
(z-z^2)^k = \sum_{i=0}^{k} \binom{k}{i} z^i (-z^2)^{\,k-i} = \sum_{i=0}^{k} \binom{k}{i} (-1)^{\,k-i} z^{i + 2(k-i)}.
$$

The power of $$z$$ in each term is

$$
i + 2(k-i) = 2k - i.
$$

We need $$2k-i = 3$$. Solving for $$i$$ gives

$$
i = 2k - 3.
$$

Since $$i$$ must lie between $$0$$ and $$k$$ (inclusive), we have two conditions:

1. $$i \ge 0 \Rightarrow 2k-3 \ge 0 \Rightarrow k \ge \frac{3}{2}$$, so $$k \ge 2$$,
2. $$i \le k \Rightarrow 2k-3 \le k \Rightarrow k \le 3$$.

Thus, the only possible values for $$k$$ are $$2$$ and $$3$$.

**Case 1: $$k=2$$**

For $$k=2$$, we have

$$
i = 2(2)-3 = 1.
$$

The contribution from this term is:

$$
\binom{9}{2} \binom{2}{1} (-1)^{\,2-1} = 36 \cdot 2 \cdot (-1) = -72.
$$

**Case 2: $$k=3$$**

For $$k=3$$, we have

$$
i = 2(3)-3 = 3.
$$

The contribution from this term is:

$$
\binom{9}{3} \binom{3}{3} (-1)^{\,3-3} = 84 \cdot 1 \cdot 1 = 84.
$$

Now, summing the contributions from $$k=2$$ and $$k=3$$, the total coefficient of $$z^3$$ is

$$
-72 + 84 = 12.
$$

Thus, the coefficient of $$z^3$$ in the expansion of $$(1+z-z^2)^9$$ is $$12$$.
The coefficient of $$ z^3 $$ in the expansion of $$ (1 + z - z^2)^9 $$ is **12**.

Write down, and simplify, the first 4 terms in the expansion of in ascending powers
of .
Replacing by , determine the coefficient of in the expansion of

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Write down, and simplify, the first 4 terms in the expansion of in ascending powers of . Replacing by , determine the coefficient of in the expansion of

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