This exercise covers proofs, evaluation of expressions, approximation, and expansion of non-binomial forms using the Binomial Theorem.
1. Proof: $a – b$ is a factor of $a^n – b^n$
Proof: We are asked to prove that $a^n – b^n$ is divisible by $a – b$ for a positive integer $n$.
We use the hint and write $a = (a – b) + b$.
Then, $a^n = ((a – b) + b)^n$. Let $x = (a – b)$.
$$a^n = (x + b)^n$$
Using the Binomial Theorem:
$$a^n = \sum_{r=0}^{n} C(n, r) x^{n-r} b^r$$
$$a^n = C(n, 0) x^n b^0 + C(n, 1) x^{n-1} b^1 + C(n, 2) x^{n-2} b^2 + \dots + C(n, n-1) x^1 b^{n-1} + C(n, n) x^0 b^n$$
Substitute $C(n, 0) = 1$, $C(n, n) = 1$, and $x = (a – b)$:
$$a^n = 1(a – b)^n(1) + C(n, 1) (a – b)^{n-1} b + C(n, 2) (a – b)^{n-2} b^2 + \dots + C(n, n-1) (a – b) b^{n-1} + 1(1) b^n$$
$$a^n = (a – b) \left[ (a – b)^{n-1} + C(n, 1) (a – b)^{n-2} b + \dots + C(n, n-1) b^{n-1} \right] + b^n$$
Rearranging the terms:
$$a^n – b^n = (a – b) \left[ (a – b)^{n-1} + C(n, 1) (a – b)^{n-2} b + \dots + C(n, n-1) b^{n-1} \right]$$
Since $a$ and $b$ are integers, the expression in the square brackets is also an integer.
Let $K = \left[ (a – b)^{n-1} + C(n, 1) (a – b)^{n-2} b + \dots + C(n, n-1) b^{n-1} \right]$.
$$a^n – b^n = (a – b) K$$
Since $K$ is an integer, $a^n – b^n$ is a multiple of $(a – b)$.
Thus, $\mathbf{a – b \text{ is a factor of } a^n – b^n}$.
2. Evaluate $(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6$
We use the general expansion of $(a+b)^n – (a-b)^n$. For $n=6$, this will cancel all terms with even powers of $b$.
$$(a+b)^6 – (a-b)^6 = 2[C(6, 1) a^5 b + C(6, 3) a^3 b^3 + C(6, 5) a^1 b^5]$$
Here $a = \sqrt{3}$ and $b = \sqrt{2}$. The coefficients $C(6, r)$ are $6, 20, 6$.
$$= 2\left[ 6(\sqrt{3})^5 (\sqrt{2})^1 + 20(\sqrt{3})^3 (\sqrt{2})^3 + 6(\sqrt{3})^1 (\sqrt{2})^5 \right]$$
Simplify the powers:
- $(\sqrt{3})^5 = 9\sqrt{3}$
- $(\sqrt{2})^5 = 4\sqrt{2}$
- $(\sqrt{3})^3 = 3\sqrt{3}$
- $(\sqrt{2})^3 = 2\sqrt{2}$
- $(\sqrt{3})^3 (\sqrt{2})^3 = (3\sqrt{3})(2\sqrt{2}) = 6\sqrt{6}$
Substitute the simplified powers:
$$= 2\left[ 6(9\sqrt{3})(\sqrt{2}) + 20(6\sqrt{6}) + 6(\sqrt{3})(4\sqrt{2}) \right]$$
$$= 2\left[ 54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6} \right]$$
$$= 2\left[ (54 + 120 + 24)\sqrt{6} \right]$$
$$= 2\left[ 198\sqrt{6} \right] = \mathbf{396\sqrt{6}}$$
3. Find the value of $\frac{1}{2} [(\sqrt{2} + 1)^4 + (\sqrt{2} – 1)^4]$
We use the general expansion of $(a+b)^n + (a-b)^n$. For $n=4$, this will cancel all terms with odd powers of $b$.
$$(a+b)^4 + (a-b)^4 = 2[C(4, 0) a^4 b^0 + C(4, 2) a^2 b^2 + C(4, 4) a^0 b^4]$$
$$(a+b)^4 + (a-b)^4 = 2[a^4 + 6a^2b^2 + b^4]$$
Now consider the expression $\frac{1}{2} [(\sqrt{2} + 1)^4 + (\sqrt{2} – 1)^4]$.
Here $a = \sqrt{2}$ and $b = 1$. Substitute into the simplified expression:
$$\frac{1}{2} \cdot 2 [a^4 + 6a^2b^2 + b^4] = a^4 + 6a^2b^2 + b^4$$
$$= (\sqrt{2})^4 + 6(\sqrt{2})^2 (1)^2 + (1)^4$$
$$= 4 + 6(2)(1) + 1$$
$$= 4 + 12 + 1 = \mathbf{17}$$
4. Find an approximation of $(0.99)^5$ using the first three terms
We write $0.99 = 1 – 0.01$. Here $a = 1$, $b = -0.01$, and $n = 5$.
$$(1 – 0.01)^5 = C(5, 0)(1)^5(-0.01)^0 + C(5, 1)(1)^4(-0.01)^1 + C(5, 2)(1)^3(-0.01)^2 + \dots$$
We use the first three terms ($r=0, 1, 2$):
$$\text{Approximation} \approx 1 – 5(0.01) + 10(0.0001)$$
$$\text{Approximation} \approx 1 – 0.05 + 0.0010$$
$$\text{Approximation} \approx 0.95 + 0.0010 = \mathbf{0.9510}$$
5. Expand using Binomial Theorem $\left(\frac{2}{x} – \frac{x}{2}\right)^4$
Here $a = \frac{2}{x}$, $b = -\frac{x}{2}$, and $n = 4$.
$$(a + b)^4 = C(4, 0)a^4 + C(4, 1)a^3b + C(4, 2)a^2b^2 + C(4, 3)ab^3 + C(4, 4)b^4$$
Using coefficients $C(4, r)$: $1, 4, 6, 4, 1$.
$$= 1\left(\frac{2}{x}\right)^4 + 4\left(\frac{2}{x}\right)^3\left(-\frac{x}{2}\right) + 6\left(\frac{2}{x}\right)^2\left(-\frac{x}{2}\right)^2 + 4\left(\frac{2}{x}\right)\left(-\frac{x}{2}\right)^3 + 1\left(-\frac{x}{2}\right)^4$$
$$= \frac{16}{x^4} + 4\left(\frac{8}{x^3}\right)\left(-\frac{x}{2}\right) + 6\left(\frac{4}{x^2}\right)\left(\frac{x^2}{4}\right) + 4\left(\frac{2}{x}\right)\left(-\frac{x^3}{8}\right) + \frac{x^4}{16}$$
$$= \frac{16}{x^4} – \frac{32x}{2x^3} + \frac{24x^2}{4x^2} – \frac{8x^3}{8x} + \frac{x^4}{16}$$
$$= \mathbf{\frac{16}{x^4} – \frac{16}{x^2} + 6 – x^2 + \frac{x^4}{16}}$$
6. Find the expansion of $(3x^2 – 2ax + 3a^2)^3$ using binomial theorem
This is a trinomial raised to a power, but it can be treated as a binomial raised to a power by grouping the first two terms: $A = (3x^2 – 2ax)$ and $B = 3a^2$.
$$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$$
- $A^3$ term: $(3x^2 – 2ax)^3$$$= (3x^2)^3 – 3(3x^2)^2(2ax) + 3(3x^2)(2ax)^2 – (2ax)^3$$$$= 27x^6 – 3(9x^4)(2ax) + 3(3x^2)(4a^2x^2) – 8a^3x^3$$$$= 27x^6 – 54ax^5 + 36a^2x^4 – 8a^3x^3 \quad (*1)$$
- $3A^2B$ term: $3(3x^2 – 2ax)^2 (3a^2) = 9a^2 (3x^2 – 2ax)^2$$$= 9a^2 [(3x^2)^2 – 2(3x^2)(2ax) + (2ax)^2]$$$$= 9a^2 [9x^4 – 12ax^3 + 4a^2x^2]$$$$= 81a^2x^4 – 108a^3x^3 + 36a^4x^2 \quad (*2)$$
- $3AB^2$ term: $3(3x^2 – 2ax) (3a^2)^2 = 3(3x^2 – 2ax) (9a^4)$$$= 27a^4 (3x^2 – 2ax)$$$$= 81a^4x^2 – 54a^5x \quad (*3)$$
- $B^3$ term: $(3a^2)^3 = 27a^6 \quad (*4)$$
Summing the terms (by descending power of $x$):
| Power of x | Coeff. from (*1) | Coeff. from (*2) | Coeff. from (*3) | Coeff. from (*4) | Total Coeff. |
| $x^6$ | $27$ | $0$ | $0$ | $0$ | $\mathbf{27}$ |
| $x^5$ | $-54a$ | $0$ | $0$ | $0$ | $\mathbf{-54a}$ |
| $x^4$ | $36a^2$ | $81a^2$ | $0$ | $0$ | $\mathbf{117a^2}$ |
| $x^3$ | $-8a^3$ | $-108a^3$ | $0$ | $0$ | $\mathbf{-116a^3}$ |
| $x^2$ | $0$ | $36a^4$ | $81a^4$ | $0$ | $\mathbf{117a^4}$ |
| $x^1$ | $0$ | $0$ | $-54a^5$ | $0$ | $\mathbf{-54a^5}$ |
| $x^0$ | $0$ | $0$ | $0$ | $27a^6$ | $\mathbf{27a^6}$ |
$$\text{Expansion} = \mathbf{27x^6 – 54ax^5 + 117a^2x^4 – 116a^3x^3 + 117a^4x^2 – 54a^5x + 27a^6}$$