Question: Expand $\left(2 - \tfrac{x}{3}\right)^5$ to the first 3 terms in ascending powers of $x$.

Step 1. Use $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ with $a = 2$, $b = -\tfrac{x}{3}$, $n = 5$.

Step 2. Write out $r = 0, 1, 2$:

$$\begin{aligned}r = 0: \;&\binom{5}{0}(2)^5 = 32 \\ r = 1: \;&\binom{5}{1}(2)^4\!\left(-\tfrac{x}{3}\right) = 5 \cdot 16 \cdot \left(-\tfrac{x}{3}\right) = -\tfrac{80}{3}x \\ r = 2: \;&\binom{5}{2}(2)^3\!\left(-\tfrac{x}{3}\right)^{\!2} = 10 \cdot 8 \cdot \tfrac{x^2}{9} = \tfrac{80}{9}x^2\end{aligned}$$

Answer:

$$\left(2 - \tfrac{x}{3}\right)^5 = 32 - \tfrac{80}{3}x + \tfrac{80}{9}x^2 + \cdots$$

⚠ Watch out: Track the sign of $b$ carefully. $\left(-\tfrac{x}{3}\right)^2$ flips back to positive, but $\left(-\tfrac{x}{3}\right)^3$ stays negative. Each $r$ alternates.

Question: Find the first 4 terms in ascending powers of $x$ in the expansion of $(3 + 2x)^6$.

Step 1. Set $a = 3$, $b = 2x$, $n = 6$. Apply $(a + b)^n = \sum \binom{n}{r} a^{n-r} b^r$.

Step 2. Compute $r = 0, 1, 2, 3$:

$$\begin{aligned}r = 0: \;&\binom{6}{0}(3)^6 = 729 \\ r = 1: \;&\binom{6}{1}(3)^5(2x) = 6 \cdot 243 \cdot 2x = 2916x \\ r = 2: \;&\binom{6}{2}(3)^4(2x)^2 = 15 \cdot 81 \cdot 4x^2 = 4860x^2 \\ r = 3: \;&\binom{6}{3}(3)^3(2x)^3 = 20 \cdot 27 \cdot 8x^3 = 4320x^3\end{aligned}$$

Answer:

$$(3 + 2x)^6 = 729 + 2916x + 4860x^2 + 4320x^3 + \cdots$$

Tip: Compute the powers of $a$, $b$, and the binomial coefficient SEPARATELY before multiplying. Fewer arithmetic slips that way.

Question: Find the term independent of $x$ in the expansion of $\left(2x^2 + \tfrac{3}{x}\right)^9$.

Step 1. General term:

$$T_{r+1} = \binom{9}{r}(2x^2)^{9-r}\!\left(\tfrac{3}{x}\right)^{\!r} = \binom{9}{r} 2^{9-r} 3^r \cdot x^{2(9-r) - r}$$

Power of $x$: $18 - 3r$.

Step 2. Term independent of $x$ requires power $= 0$:

$$18 - 3r = 0 \;\Rightarrow\; r = 6$$

Step 3. Substitute $r = 6$:

$$T_7 = \binom{9}{6} \cdot 2^3 \cdot 3^6 = 84 \cdot 8 \cdot 729 = 489{,}888$$

Answer: The term independent of $x$ is $489{,}888$.

Tip: "Term independent of $x$" = power of $x$ is zero. Set the general power $= 0$, solve for $r$, substitute back.

Question: Find the coefficient of $x^3$ in the expansion of $\left(x - \tfrac{2}{x^2}\right)^9$.

Step 1. General term:

$$T_{r+1} = \binom{9}{r} x^{9-r}\!\left(-\tfrac{2}{x^2}\right)^{\!r} = \binom{9}{r}(-2)^r \cdot x^{9 - r - 2r} = \binom{9}{r}(-2)^r \cdot x^{9 - 3r}$$

Step 2. Set $9 - 3r = 3 \Rightarrow r = 2$.

Step 3. Substitute $r = 2$:

$$T_3 = \binom{9}{2}(-2)^2 = 36 \cdot 4 = 144$$

Answer: Coefficient of $x^3$ is $144$.

⚠ Watch out: The sign of $(-2)^r$ matters. Here $r = 2$ (even) so the contribution is positive. If $r$ had been odd, the coefficient would be negative.

Question: Explain why there is no term independent of $x$ in the expansion of $\left(\tfrac{x^2}{3} - \tfrac{2}{x^3}\right)^9$.

Step 1. General term:

$$T_{r+1} = \binom{9}{r}\!\left(\tfrac{x^2}{3}\right)^{\!9-r}\!\left(-\tfrac{2}{x^3}\right)^{\!r}$$

Power of $x$: $2(9 - r) - 3r = 18 - 5r$.

Step 2. For a term independent of $x$, set the power $= 0$:

$$18 - 5r = 0 \;\Rightarrow\; r = \tfrac{18}{5}$$

Step 3. Since $r$ must be a non-negative integer in $[0, 9]$ and $\tfrac{18}{5} = 3.6$ is not an integer, no such $r$ exists.

Conclusion: There is no term independent of $x$ in this expansion.

Why this works: $r$ MUST be a whole number for $T_{r+1}$ to be a valid term. If solving "general power $=$ target" gives a non-integer $r$, that target power doesn't appear at all.

Question: Show that every term in the expansion of $\left(x^3 + \tfrac{1}{x}\right)^5$ has an odd power of $x$.

Step 1. General term:

$$T_{r+1} = \binom{5}{r}(x^3)^{5-r}\!\left(\tfrac{1}{x}\right)^{\!r} = \binom{5}{r} \cdot x^{3(5-r) - r} = \binom{5}{r} \cdot x^{15 - 4r}$$

Step 2. Power of $x$ is $15 - 4r$. As $r$ ranges over $0, 1, 2, 3, 4, 5$:

$r$	$15 - 4r$
0	15
1	11
2	7
3	3
4	$-1$
5	$-5$

All odd.

Step 3. Algebraic reason: $15$ is odd and $4r$ is always even, so $15 - 4r$ is always odd.

Conclusion: Every term has an odd power of $x$.

Tip: To classify the parity of powers, look at the general-power formula. "Odd constant $-$ even multiple of $r$" $\Rightarrow$ all powers odd. Same idea works for "all even."

Question: In the expansion of $(2 + ax)^n$, the coefficients of $x^2$ and $x^3$ are in the ratio $1 : 5$. Given $a = 5$, find $n$.

Step 1. General term coefficients:

• Coefficient of $x^2$: $\binom{n}{2} 2^{n-2} a^2$
• Coefficient of $x^3$: $\binom{n}{3} 2^{n-3} a^3$

Step 2. Set up the ratio:

$$\dfrac{\binom{n}{2} 2^{n-2} a^2}{\binom{n}{3} 2^{n-3} a^3} = \dfrac{1}{5}$$

Simplify: $\dfrac{2^{n-2}}{2^{n-3}} = 2$ and $\dfrac{a^2}{a^3} = \dfrac{1}{a}$, so:

$$\dfrac{2\binom{n}{2}}{a\binom{n}{3}} = \dfrac{1}{5}$$

Step 3. Using $\binom{n}{2} = \tfrac{n(n-1)}{2}$ and $\binom{n}{3} = \tfrac{n(n-1)(n-2)}{6}$:

$$\dfrac{2 \cdot \tfrac{n(n-1)}{2}}{a \cdot \tfrac{n(n-1)(n-2)}{6}} = \dfrac{1}{5} \;\Rightarrow\; \dfrac{6}{a(n-2)} = \dfrac{1}{5} \;\Rightarrow\; a(n-2) = 30$$

With $a = 5$: $n - 2 = 6 \Rightarrow n = 8$.

Tip: Cancel $\binom{n}{2} / \binom{n}{3}$ and the matching powers of $2$ and $a$ BEFORE expanding factorials — much less arithmetic.

Question: The first three terms of $(1 + kx)^n$ in ascending powers of $x$ are $1$, $30x$, and $405x^2$. Find $n$ and $k$.

Step 1. Expand:

$$(1 + kx)^n = 1 + n(kx) + \binom{n}{2}(kx)^2 + \cdots = 1 + nk\,x + \tfrac{n(n-1)k^2}{2}\,x^2 + \cdots$$

Step 2. Match coefficients:

• $x$: $\;nk = 30 \quad (1)$
• $x^2$: $\;\tfrac{n(n-1)k^2}{2} = 405 \quad (2)$

Step 3. From (1), $k = \tfrac{30}{n}$. Substitute into (2):

$$\tfrac{n(n-1) \cdot \tfrac{900}{n^2}}{2} = 405 \;\Rightarrow\; \tfrac{900(n-1)}{2n} = 405$$ $$900(n-1) = 810n \;\Rightarrow\; 90n = 900 \;\Rightarrow\; n = 10$$

Then $k = \tfrac{30}{10} = 3$.

Answer: $n = 10$, $k = 3$.

Tip: When two equations contain $n$ and another unknown, isolate the easier one first (linear in $k$ here), substitute, and the second becomes a single-variable equation in $n$.

Question: The coefficient of $x^3$ in the expansion of $(10 + kx)\!\left(2 - \tfrac{x}{2}\right)^{\!8}$ is zero. Find $k$.

Step 1. Find the relevant coefficients from $\left(2 - \tfrac{x}{2}\right)^8$:

$$\text{Coeff of }x^2: \;\binom{8}{2}(2)^6\!\left(-\tfrac{1}{2}\right)^{\!2} = 28 \cdot 64 \cdot \tfrac{1}{4} = 448$$ $$\text{Coeff of }x^3: \;\binom{8}{3}(2)^5\!\left(-\tfrac{1}{2}\right)^{\!3} = 56 \cdot 32 \cdot \left(-\tfrac{1}{8}\right) = -224$$

Step 2. When multiplying by $(10 + kx)$:

$$\text{coeff of }x^3 \text{ in product} = 10 \cdot (\text{coeff }x^3) + k \cdot (\text{coeff }x^2)$$ $$= 10(-224) + k(448) = -2240 + 448k$$

Step 3. Set to zero:

$$-2240 + 448k = 0 \;\Rightarrow\; k = 5$$

Tip: When multiplying $(A + Bx)$ by a binomial, the coefficient of $x^k$ in the product is $A \cdot (\text{coeff }x^k) + B \cdot (\text{coeff }x^{k-1})$. Pull only the two coefficients you need — don't expand fully.

Question: Find the constant term in the expansion of $(2x - 1)\!\left(\tfrac{1}{x^2} - 3x\right)^{\!6}$.

Step 1. General term in $\left(\tfrac{1}{x^2} - 3x\right)^6$:

$$T_{r+1} = \binom{6}{r}\!\left(\tfrac{1}{x^2}\right)^{\!6-r}(-3x)^r = \binom{6}{r}(-3)^r \cdot x^{-2(6-r) + r} = \binom{6}{r}(-3)^r \cdot x^{3r - 12}$$

Step 2. Multiplying by $(2x - 1)$ shifts the powers:

• From $2x$: contributes powers $3r - 12 + 1 = 3r - 11$
• From $-1$: contributes powers $3r - 12$

For a constant term (power $= 0$):

• $3r - 11 = 0 \Rightarrow r = \tfrac{11}{3}$ — not integer, no contribution.
• $3r - 12 = 0 \Rightarrow r = 4$ ✓

Step 3. Constant contribution from $r = 4$:

$$-1 \cdot \binom{6}{4}(-3)^4 = -1 \cdot 15 \cdot 81 = -1215$$

Answer: Constant term is $-1215$.

Watch out: Both parts of the pre-factor can contribute to the constant — check each separately by solving for $r$.

Question: Use the first four terms of $\left(2 + \tfrac{x}{2}\right)^8$ to estimate $(2.0025)^8$ to 3 decimal places.

Step 1. Expand to four terms:

$$\begin{aligned}\binom{8}{0}\cdot 2^8 &= 256 \\ \binom{8}{1}\cdot 2^7 \cdot \tfrac{x}{2} &= 512x \\ \binom{8}{2}\cdot 2^6 \cdot \tfrac{x^2}{4} &= 448 x^2 \\ \binom{8}{3}\cdot 2^5 \cdot \tfrac{x^3}{8} &= 224 x^3\end{aligned}$$

So $\left(2 + \tfrac{x}{2}\right)^8 = 256 + 512x + 448x^2 + 224x^3 + \cdots$

Step 2. Choose $x$ so that $2 + \tfrac{x}{2} = 2.0025$:

$$\tfrac{x}{2} = 0.0025 \;\Rightarrow\; x = 0.005$$

Step 3. Substitute:

$$\begin{aligned}(2.0025)^8 &\approx 256 + 512(0.005) + 448(0.005)^2 + 224(0.005)^3 \\ &= 256 + 2.56 + 0.0112 + 0.000028 \\ &\approx 258.571\end{aligned}$$

Tip: Smaller $|x|$ = more accurate truncation. Each subsequent term is roughly $|x|$ smaller, so $|x| = 0.005$ gives ~3-4 decimals of accuracy from just 3-4 terms.

Question: Use the first three terms of $(1 + 2x)^{10}$ to approximate $(1.02)^{10}$ to 4 decimal places.

Step 1. Expand to three terms:

$$(1 + 2x)^{10} = 1 + 10(2x) + \binom{10}{2}(2x)^2 + \cdots = 1 + 20x + 180x^2 + \cdots$$

Step 2. Choose $x$ so that $1 + 2x = 1.02$:

$$2x = 0.02 \;\Rightarrow\; x = 0.01$$

Step 3. Substitute:

$$\begin{aligned}(1.02)^{10} &\approx 1 + 20(0.01) + 180(0.01)^2 \\ &= 1 + 0.2 + 0.018 \\ &= 1.218\end{aligned}$$

To 4 decimal places: $(1.02)^{10} \approx 1.2180$.

Improve accuracy: Add the next term, $\binom{10}{3}(2x)^3 = 120 \cdot 8 \cdot 10^{-6} = 0.00096$, giving $1.21896$. The true value is $1.21899\ldots$, so 4-term truncation is accurate to 4 d.p.

Question: Show that $(2 + x)^6$ leaves a remainder of $1$ when divided by $(1 + x)$.

Step 1. By the remainder theorem, the remainder when a polynomial $P(x)$ is divided by $(1 + x)$ equals $P(-1)$.

Step 2. Compute $(2 + x)^6$ at $x = -1$:

$$P(-1) = (2 + (-1))^6 = 1^6 = 1$$

Conclusion: The remainder is $1$. ✓

Why this works: No need to expand $(2 + x)^6$ at all. The remainder theorem skips straight to the answer with one substitution.

Tip: Many "show that ... remainder ... " problems disguised as binomial questions are really one-line applications of the remainder theorem. Spot the divisor of form $(x \pm a)$ and substitute.

Question: Use the substitution $z = 1 + x$ to find the coefficient of $x^2$ in the expansion of $[1 + (1 + x)]^6$.

Step 1. Let $z = 1 + x$. Then $[1 + (1+x)]^6 = (1 + z)^6$.

Expand:

$$(1 + z)^6 = 1 + 6z + 15z^2 + 20z^3 + 15z^4 + 6z^5 + z^6$$

Step 2. Substitute back $z = 1 + x$. We need the coefficient of $x^2$. Each $z^k = (1 + x)^k$ contributes $\binom{k}{2}$ to the $x^2$ coefficient:

$$\text{coeff of }x^2 = 6 \cdot \binom{1}{2} + 15 \cdot \binom{2}{2} + 20 \cdot \binom{3}{2} + 15 \cdot \binom{4}{2} + 6 \cdot \binom{5}{2} + 1 \cdot \binom{6}{2}$$

Note $\binom{1}{2} = 0$. Compute:

$$= 0 + 15 \cdot 1 + 20 \cdot 3 + 15 \cdot 6 + 6 \cdot 10 + 1 \cdot 15 = 15 + 60 + 90 + 60 + 15 = 240$$

Sanity check: $[1 + (1 + x)]^6 = (2 + x)^6$, and the coefficient of $x^2$ in $(2 + x)^6$ is $\binom{6}{2} \cdot 2^4 = 15 \cdot 16 = 240$ ✓.

Why substitute: $z = 1 + x$ converts a "binomial inside a binomial" into a single binomial, which can be easier to extract terms from when the structure permits.