Exercise 10.2: Binomial Expansion
1. The coefficients of expansion $(x + y)^5$ are:
Solution: According to the Binomial Theorem, expanding $(x + y)^5$ gives the coefficients:
$\binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \binom{5}{3}, \binom{5}{4}, \binom{5}{5}$
Calculating the values:
$\binom{5}{0} = 1$
$\binom{5}{1} = 5$
$\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$
$\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$
$\binom{5}{4} = 5$
$\binom{5}{5} = 1$
Therefore, the coefficients are: $1, 5, 10, 10, 5, 1$
Answer: b) $1, 5, 10, 10, 5, 1$
2. What is the $x$ free term in the expansion of $\left(x^2 + \frac{1}{x^2}\right)^4$?
Solution: Expanding using the binomial theorem we get,
$\left(x^2 + \frac{1}{x^2}\right)^4 = (x^2)^4 + \binom{4}{1}(x^2)^3\left(\frac{1}{x^2}\right) + \binom{4}{2}(x^2)^2\left(\frac{1}{x^2}\right)^2 + \binom{4}{3}(x^2)\left(\frac{1}{x^2}\right)^3 + \left(\frac{1}{x^2}\right)^4$
$= x^8 + 4x^6\cdot\frac{1}{x^2} + 6x^4\cdot\frac{1}{x^4} + 4x^2\cdot\frac{1}{x^6} + \frac{1}{x^8}$
$= x^8 + 4x^4 + 6 + \frac{4}{x^4} + \frac{1}{x^8}$
Here, the term independent of $x$ (constant term) is $6$.
Answer: b) $6$
3. Ordering the coefficients of the expansion $(x + y)^4$, we get—
Solution: Ordering the powers from $n = 0$ to $n = 4$ using Pascal’s triangle rule:
$n = 0 \to 1$
$n = 1 \to 1 \quad 1$
$n = 2 \to 1 \quad 2 \quad 1$
$n = 3 \to 1 \quad 3 \quad 3 \quad 1$
$n = 4 \to 1 \quad 4 \quad 6 \quad 4 \quad 1$
This sequence exactly matches option ‘b’.
Answer: b)
$1$
$1 \quad 1$
$1 \quad 2 \quad 1$
$1 \quad 3 \quad 3 \quad 1$
$1 \quad 4 \quad 6 \quad 4 \quad 1$
4. Expand each of these:
1) $(2 + x^2)^5$
Solution: Applying the binomial theorem we get,
$(2 + x^2)^5$
$= 2^5 + \binom{5}{1} \cdot 2^4 \cdot (x^2) + \binom{5}{2} \cdot 2^3 \cdot (x^2)^2 + \binom{5}{3} \cdot 2^2 \cdot (x^2)^3 + \binom{5}{4} \cdot 2^1 \cdot (x^2)^4 + (x^2)^5$
$= 32 + (5 \times 16 \times x^2) + (10 \times 8 \times x^4) + (10 \times 4 \times x^6) + (5 \times 2 \times x^8) + x^{10}$
$= 32 + 80x^2 + 80x^4 + 40x^6 + 10x^8 + x^{10}$
Answer: $32 + 80x^2 + 80x^4 + 40x^6 + 10x^8 + x^{10}$
2) $\left(2 – \frac{1}{2x}\right)^6$
Solution: Applying the binomial theorem we get,
$\left(2 – \frac{1}{2x}\right)^6$
$= 2^6 + \binom{6}{1} \cdot 2^5 \cdot \left(-\frac{1}{2x}\right) + \binom{6}{2} \cdot 2^4 \cdot \left(-\frac{1}{2x}\right)^2 + \binom{6}{3} \cdot 2^3 \cdot \left(-\frac{1}{2x}\right)^3 + \binom{6}{4} \cdot 2^2 \cdot \left(-\frac{1}{2x}\right)^4 + \binom{6}{5} \cdot 2^1 \cdot \left(-\frac{1}{2x}\right)^5 + \left(-\frac{1}{2x}\right)^6$
Evaluating and simplifying the values, we get
$= 64 + 6 \cdot 32 \cdot \left(-\frac{1}{2x}\right) + 15 \cdot 16 \cdot \left(\frac{1}{4x^2}\right) + 20 \cdot 8 \cdot \left(-\frac{1}{8x^3}\right) + 15 \cdot 4 \cdot \left(\frac{1}{16x^4}\right) + 6 \cdot 2 \cdot \left(-\frac{1}{32x^5}\right) + \frac{1}{64x^6}$
$= 64 – \frac{192}{2x} + \frac{240}{4x^2} – \frac{160}{8x^3} + \frac{60}{16x^4} – \frac{12}{32x^5} + \frac{1}{64x^6}$
$= 64 – \frac{96}{x} + \frac{60}{x^2} – \frac{20}{x^3} + \frac{15}{4x^4} – \frac{3}{8x^5} + \frac{1}{64x^6}$
Answer: $64 – \frac{96}{x} + \frac{60}{x^2} – \frac{20}{x^3} + \frac{15}{4x^4} – \frac{3}{8x^5} + \frac{1}{64x^6}$
5. Determine the first four terms of these expansions:
1) $(2 + 3x)^6$
Solution: Applying the binomial theorem, we expand up to the first four terms-
$(2 + 3x)^6 = 2^6 + \binom{6}{1} 2^5 (3x) + \binom{6}{2} 2^4 (3x)^2 + \binom{6}{3} 2^3 (3x)^3 + \cdots$
$= 64 + 6 \times 32 \times 3x + 15 \times 16 \times 9x^2 + 20 \times 8 \times 27x^3 + \cdots$
$= 64 + 576x + 2160x^2 + 4320x^3 + \cdots$
Answer: The first four terms are: $64, 576x, 2160x^2, 4320x^3$
2) $\left(4 – \frac{1}{2x}\right)^5$
Solution:
Applying the binomial theorem, we expand up to the first four terms-
$\left(4 – \frac{1}{2x}\right)^5 = 4^5 + \binom{5}{1} 4^4 \left(-\frac{1}{2x}\right) + \binom{5}{2} 4^3 \left(-\frac{1}{2x}\right)^2 + \binom{5}{3} 4^2 \left(-\frac{1}{2x}\right)^3 + \cdots$
$= 1024 + 5 \times 256 \times \left(-\frac{1}{2x}\right) + 10 \times 64 \times \left(\frac{1}{4x^2}\right) + 10 \times 16 \times \left(-\frac{1}{8x^3}\right) + \cdots$
$= 1024 – \frac{640}{x} + \frac{160}{x^2} – \frac{20}{x^3} + \cdots$
Answer: The first four terms are: $1024, -\frac{640}{x}, \frac{160}{x^2}, -\frac{20}{x^3}$
If $\left(p – \frac{1}{2}x\right)^6 = r – 96x + sx^2 + \cdots$, determine $p, r,$ and $s$.
Solution: Expanding the left side, we get
$\left(p – \frac{1}{2}x\right)^6 = p^6 + \binom{6}{1} p^5 \left(-\frac{1}{2}x\right) + \binom{6}{2} p^4 \left(-\frac{1}{2}x\right)^2 + \cdots$
$= p^6 – 3p^5 x + \frac{15}{4}p^4 x^2 + \cdots$
According to the question,
$p^6 – 3p^5 x + \frac{15}{4}p^4 x^2 + \cdots = r – 96x + sx^2 + \cdots$
Equating coefficients from both sides. we have ,
$r = p^6$
$-3p^5 = -96 \implies p^5 = 32 \implies p^5 = 2^5 \implies p = 2$
$s = \frac{15}{4}p^4$
Substituting $p = 2$, we get
$r = 2^6 = 64$
$s = \frac{15}{4} \times 2^4 = \frac{15}{4} \times 16 = 60$
Answer: $p = 2, r = 64, s = 60$
7. Expand $\left(2 + \frac{x}{4}\right)^6$ up to $x^3$ in ascending powers of $x$. Find the approximate value of $(1.9975)^6$ up to four decimal places.
Solution:
$\left(2 + \frac{x}{4}\right)^6 = 2^6 + \binom{6}{1} 2^5 \left(\frac{x}{4}\right) + \binom{6}{2} 2^4 \left(\frac{x}{4}\right)^2 + \binom{6}{3} 2^3 \left(\frac{x}{4}\right)^3 + \cdots$
$= 64 + 6 \times 32 \times \frac{x}{4} + 15 \times 16 \times \frac{x^2}{16} + 20 \times 8 \times \frac{x^3}{64} + \cdots$
$= 64 + 48x + 15x^2 + \frac{5}{2}x^3 + \cdots$ …….. (1)
Now,
$2 + \frac{x}{4} = 1.9975$
or, $\frac{x}{4} = 1.9975 – 2 = -0.0025$
or, $x = 4 \times (-0.0025) = -0.01$
Substituting $x = -0.01$ into equation (1):
$(1.9975)^6 \approx 64 + 48(-0.01) + 15(-0.01)^2 + 2.5(-0.01)^3$
$= 64 – 0.48 + 15(0.0001) + 2.5(-0.000001)$
$= 64 – 0.48 + 0.0015 – 0.0000025$
$= 63.5214975 \approx 63.5215$
Answer: $63.5215$ (up to four decimal places).
8. Using the binomial theorem, find the value of $(1.99)^5$ up to four decimal places.
Solution:
$(1.99)^5 = (2 – 0.01)^5$
Expanding using the binomial theorem, we get
$(2 – 0.01)^5 = 2^5 + \binom{5}{1} 2^4 (-0.01) + \binom{5}{2} 2^3 (-0.01)^2 + \binom{5}{3} 2^2 (-0.01)^3 + \binom{5}{4} 2^1 (-0.01)^4 + (-0.01)^5$
$= 32 + 5 \times 16 \times (-0.01) + 10 \times 8 \times (0.0001) + 10 \times 4 \times (-0.000001) + \cdots$
$= 32 – 0.8 + 0.008 – 0.00004 + \cdots$
$= 31.20796 \approx 31.2080$
Answer: $31.2080$ (up to four decimal places).
9. In the expansion of $\left(1 + \frac{x}{4}\right)^n$, coefficient of 3rd term is the double of the coefficient of the 4th term. Find the value of $n$. Also, determine the number of terms and the middle term of the expansion.
Solution:
The expanded form of $\left(1 + \frac{x}{4}\right)^n$:
Third term, $T_3 = \binom{n}{2} \left(\frac{x}{4}\right)^2 = \frac{n(n-1)}{2} \cdot \frac{x^2}{16} \implies \text{Coefficient} = \frac{n(n-1)}{32}$
Fourth term, $T_4 = \binom{n}{3} \left(\frac{x}{4}\right)^3 = \frac{n(n-1)(n-2)}{6} \cdot \frac{x^3}{64} \implies \text{Coefficient} = \frac{n(n-1)(n-2)}{384}$
According to the question,
$\frac{n(n-1)}{32} = 2 \times \frac{n(n-1)(n-2)}{384}$
Dividing both sides by $n(n-1)$ (where $n \neq 0, 1$):
$\frac{1}{32} = \frac{n-2}{192}$
or, $32(n – 2) = 192$
or, $n – 2 = \frac{192}{32}$
or, $n – 2 = 6 \implies n = 8$
Number of terms: $n + 1 = 8 + 1 = 9$.
Middle term: Since the number of terms is 9 (odd), the middle term is $\left(\frac{8}{2} + 1\right) = 5\text{th}$ term.
Middle term ($T_5$) $= \binom{8}{4} \left(\frac{x}{4}\right)^4 = 70 \times \frac{x^4}{256} = \frac{35}{128}x^4$
Answer: $n = 8$, Number of terms $= 9$, Middle term $= \frac{35}{128}x^4$
10. In the expansion of $\left(x^2 + \frac{k}{x}\right)^6$ coefficient of $x^3$ is $160$, find the value of $k$.
Solution:
General term $T_{r+1} = \binom{6}{r} (x^2)^{6-r} \left(\frac{k}{x}\right)^r = \binom{6}{r} x^{12-2r} \cdot k^r \cdot x^{-r} = \binom{6}{r} k^r x^{12-3r}$
For the coefficient of $x^3$:
$12 – 3r = 3 \implies 3r = 9 \implies r = 3$
Therefore, the coefficient of $x^3 = \binom{6}{3} k^3 = 20k^3$
According to the question,
$20k^3 = 160$
or, $k^3 = 8 \implies k^3 = 2^3 \implies k = 2$
Answer: $k = 2$
11. $(A + Bx)^n$ is an algebraic expression.
1)
If $A = 1, B = 2$, and $n = 5$, determine the expansion of the expression using Pascal’s Triangle.
Solution:
The expression becomes $(1 + 2x)^5$
For $n = 5$, the coefficients from Pascal’s triangle are: $1, 5, 10, 10, 5, 1$
Therefore,
$(1 + 2x)^5 = 1(1) + 5(1)(2x) + 10(1)(2x)^2 + 10(1)(2x)^3 + 5(1)(2x)^4 + 1(2x)^5$
$= 1 + 10x + 10(4x^2) + 10(8x^3) + 5(16x^4) + 32x^5$
$= 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5$
Answer: $1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5$
2)
If $B = 3$ and $n = 7$, in the expansion of the expression,coefficient of $x^4$ is $22680$. Determine $A$.
Solution:
The expression is $(A + 3x)^7$
The term containing $x^4$ is $T_5 = \binom{7}{4} A^{7-4} (3x)^4 = 35 \cdot A^3 \cdot 81x^4 = 2835 A^3 x^4$
Coefficient of $x^4 = 2835 A^3$
According to the question,
$2835 A^3 = 22680$
or, $A^3 = \frac{22680}{2835} = 8$
or, $A^3 = 2^3 \implies A = 2$
Answer: $A = 2$
3)
If $A = 2$ and $B = 1$, then the coefficients of 5th and 6th terms of the expansion are same. Determine the value of $n$.
Solution:
The expression is $(2 + x)^n$
Fifth term, $T_5 = \binom{n}{4} 2^{n-4} x^4 \implies \text{Coefficient} = \binom{n}{4} 2^{n-4}$
Sixth term, $T_6 = \binom{n}{5} 2^{n-5} x^5 \implies \text{Coefficient} = \binom{n}{5} 2^{n-5}$
According to the question,
$\binom{n}{4} 2^{n-4} = \binom{n}{5} 2^{n-5}$
or, $\frac{\binom{n}{4}}{\binom{n}{5}} = \frac{2^{n-5}}{2^{n-4}}$
or, $\frac{\frac{n!}{4!(n-4)!}}{\frac{n!}{5!(n-5)!}} = 2^{(n-5)-(n-4)}$
or, $\frac{5!(n-5)!}{4!(n-4)!} = 2^{-1}$
or, $\frac{5 \times (n-5)!}{(n-4)(n-5)!} = \frac{1}{2}$
or, $\frac{5}{n-4} = \frac{1}{2} \implies n – 4 = 10 \implies n = 14$
Answer: $n = 14$
12. If $a_1, a_2, a_3, a_4$ are the coefficients of four consecutive terms in the expansion of $(1 + x)^n$, then prove that
$\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2a_2}{a_2 + a_3}$
Proof:
Let the four consecutive terms in the expansion of $(1 + x)^n$ be the $r\text{th}$, $(r+1)\text{th}$, $(r+2)\text{th}$, and $(r+3)\text{th}$ terms respectively.
Therefore,
$a_1 = \binom{n}{r-1}$
$a_2 = \binom{n}{r}$
$a_3 = \binom{n}{r+1}$
$a_4 = \binom{n}{r+2}$
We know that $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$
So,
$a_1 + a_2 = \binom{n+1}{r}$
$a_2 + a_3 = \binom{n+1}{r+1}$
$a_3 + a_4 = \binom{n+1}{r+2}$
Now, evaluating the parts on the left side:
$\frac{a_1}{a_1 + a_2} = \frac{\binom{n}{r-1}}{\binom{n+1}{r}} = \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{(n+1)!}{r!(n-r+1)!}} = \frac{r}{n+1}$
$\frac{a_3}{a_3 + a_4} = \frac{\binom{n}{r+1}}{\binom{n+1}{r+2}} = \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{(n+1)!}{(r+2)!(n-r-1)!}} = \frac{r+2}{n+1}$
Left Hand Side $= \frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{r}{n+1} + \frac{r+2}{n+1} = \frac{2r + 2}{n+1} = \frac{2(r+1)}{n+1}$
Now for the Right Hand Side:
$\frac{2a_2}{a_2 + a_3} = 2 \times \frac{\binom{n}{r}}{\binom{n+1}{r+1}} = 2 \times \frac{\frac{n!}{r!(n-r)!}}{\frac{(n+1)!}{(r+1)!(n-r)!}} = 2 \times \frac{r+1}{n+1} = \frac{2(r+1)}{n+1}$
Therefore, $\text{Left Hand Side} = \text{Right Hand Side}$ (Proved)
Which one is bigger? $99^{50} + 100^{50}$ or $101^{50}$?
Solution:
We will evaluate the value of $101^{50} – 99^{50}$ using the binomial theorem and compare it with $100^{50}$.
We know,
$101^{50} = (100 + 1)^{50}$
$99^{50} = (100 – 1)^{50}$
Applying binomial expansion:
$(100 + 1)^{50} = 100^{50} + \binom{50}{1} 100^{49} + \binom{50}{2} 100^{48} + \binom{50}{3} 100^{47} + \cdots + 1$
$(100 – 1)^{50} = 100^{50} – \binom{50}{1} 100^{49} + \binom{50}{2} 100^{48} – \binom{50}{3} 100^{47} + \cdots + 1$
Subtracting the 2nd equation from the 1st equation:
$101^{50} – 99^{50} = 2 \cdot \left[ \binom{50}{1} 100^{49} + \binom{50}{3} 100^{47} + \binom{50}{5} 100^{45} + \cdots \right]$
Here the 1st term is:
$2 \cdot \binom{50}{1} 100^{49} = 2 \cdot 50 \cdot 100^{49} = 100 \cdot 100^{49} = 100^{50}$
Since all remaining terms of the series ($\binom{50}{3} 100^{47}$, $\binom{50}{5} 100^{45}$, etc.) are positive numbers, we have:
$101^{50} – 99^{50} > 100^{50}$
Rearranging:
$101^{50} > 99^{50} + 100^{50}$
Answer: $101^{50}$ is larger.
Sample Questions (Multiple Choice)
1. In the expansion of $(1 – x)\left(1 + \frac{x}{2}\right)^8$, the coefficient of $x$ is—
Solution:
First, expand $\left(1 + \frac{x}{2}\right)^8$:
$\left(1 + \frac{x}{2}\right)^8 = 1 + \binom{8}{1}\left(\frac{x}{2}\right) + \cdots = 1 + 8 \cdot \frac{x}{2} + \cdots = 1 + 4x + \cdots$
Now,
$(1 – x)\left(1 + \frac{x}{2}\right)^8 = (1 – x)(1 + 4x + \cdots)$
Multiplying to find terms containing $x$:
$= 1 \cdot (4x) + (-x) \cdot 1 = 4x – x = 3x$
Therefore, coefficient of $x = 3$
Answer: c) $3$
2. In the expansion of $(1 + 2x + x^2)^3$—
(i) Number of terms is $7$
(ii) $2\text{nd}$ term is $6x$
(iii) Last term is $x^6$
Verification:
Simplifying the expression:
$(1 + 2x + x^2)^3 = \{(1 + x)^2\}^3 = (1 + x)^6$
(i) Number of terms: In binomial expansion $(1 + x)^n$, number of terms is $(n + 1)$. Here $n = 6$, so number of terms $= 6 + 1 = 7$. [Statement (i) is true]
(ii) $2\text{nd}$ term: $T_2 = \binom{6}{1} (1)^{6-1} (x)^1 = 6x$. [Statement (ii) is true]
(iii) Last term: $T_7 = \binom{6}{6} x^6 = x^6$. [Statement (iii) is true]
Answer: d) $i, ii \text{ and } iii$
Stem (For Questions 3 and 4):
$\left(x + \frac{1}{x}\right)^n$, where $n$ is an even number.
3. If $(r + 1)\text{th}$ term is free of $x$, what is the value of $r$?
Solution: General term,
$T_{r+1} = \binom{n}{r} (x)^{n-r} \left(\frac{1}{x}\right)^r = \binom{n}{r} x^{n-r} \cdot x^{-r} = \binom{n}{r} x^{n-2r}$
Since the term is independent of $x$ (i.e., $x^0$):
$n – 2r = 0 \implies 2r = n \implies r = \frac{n}{2}$
Answer: b) $\frac{n}{2}$
4. If $n = 4$, then which one is the fourth term?
Solution:
If $n = 4$, the expression becomes $\left(x + \frac{1}{x}\right)^4$
Fourth term ($T_4 = T_{3+1}$):
$T_4 = \binom{4}{3} (x)^{4-3} \left(\frac{1}{x}\right)^3$
$= 4 \cdot x^1 \cdot \frac{1}{x^3}$
$= 4 \cdot \frac{1}{x^2} = \frac{4}{x^2}$
Answer: d) $\frac{4}{x^2}$
Creative Questions
5. $A = (1 + x)^7$ and $B = (1 – x)^8$
a) Determine the expansion of $A$ using Pascal’s Triangle.
Solution: $A = (1 + x)^7$
For $n = 7$, Pascal’s triangle coefficients are:
$1 \quad 7 \quad 21 \quad 35 \quad 35 \quad 21 \quad 7 \quad 1$
Expansion using Pascal’s triangle coefficients:
$(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7$
b) Expand $B$ up to four terms. Use the result to find the value of $(0.99)^8$ up to four decimal places.
Solution:
$B = (1 – x)^8$
Expansion up to first four terms, we have
$(1 – x)^8 = 1 + \binom{8}{1}(-x) + \binom{8}{2}(-x)^2 + \binom{8}{3}(-x)^3 + \cdots$
$= 1 – 8x + 28x^2 – 56x^3 + \cdots$ …….. (1)
Now, assuming $(1 – x)^8 = (0.99)^8$,
$1 – x = 0.99 \implies x = 1 – 0.99 = 0.01$
Substituting $x = 0.01$ into equation (1):
$(0.99)^8 \approx 1 – 8(0.01) + 28(0.01)^2 – 56(0.01)^3$
$= 1 – 0.08 + 28(0.0001) – 56(0.000001)$
$= 1 – 0.08 + 0.0028 – 0.000056$
$= 0.922744 \approx 0.9227$
Answer: $0.9227$ (up to four decimal places)
c) Determine the coefficient of $x^7$ in the expansion of $AB$.
Solution:
$AB = (1 + x)^7 (1 – x)^8 = (1 + x)^7 (1 – x)^7 (1 – x)$
$= \{(1 + x)(1 – x)\}^7 (1 – x)$
$= (1 – x^2)^7 (1 – x)$
Expanding $(1 – x^2)^7$ using the binomial theorem:
$(1 – x^2)^7 = \binom{7}{0} – \binom{7}{1}x^2 + \binom{7}{2}x^4 – \binom{7}{3}x^6 + \binom{7}{4}x^8 – \cdots$
$= 1 – 7x^2 + 21x^4 – 35x^6 + 35x^8 – \cdots$
Therefore,
$AB = (1 – 7x^2 + 21x^4 – 35x^6 + 35x^8 – \cdots)(1 – x)$
Calculating terms containing $x^7$ from the product:
$= (-35x^6) \cdot (-x) + (35x^8) \cdot 0 = 35x^7$
Therefore, coefficient of $x^7 = 35$
Answer: $35$
6. $A = \left(2x^2 – \frac{1}{2x^3}\right)^{10}$ and $B = \left(3k – \frac{2x}{3}\right)^5$
a) Find the value of the middle term in the expansion of $A$.
Solution:
$A = \left(2x^2 – \frac{1}{2x^3}\right)^{10}$
Here $n = 10$ (even number). Number of terms $= 10 + 1 = 11$.
The middle term is $\left(\frac{10}{2} + 1\right) = 6\text{th}$ term.
$6\text{th}$ term ($T_6 = T_{5+1}$):
$T_6 = \binom{10}{5} (2x^2)^{10-5} \left(-\frac{1}{2x^3}\right)^5$
$= 252 \cdot (2x^2)^5 \cdot \left(-\frac{1}{32x^{15}}\right)$
$= 252 \cdot 32x^{10} \cdot \left(-\frac{1}{32x^{15}}\right)$
$= -252 \cdot \frac{x^{10}}{x^{15}} = -\frac{252}{x^5}$
Answer: $-\frac{252}{x^5}$
b) Determine the term free of $x$ in the expansion of $A$.
Solution:
General term $T_{r+1} = \binom{10}{r} (2x^2)^{10-r} \left(-\frac{1}{2x^3}\right)^r$
$= \binom{10}{r} 2^{10-r} x^{20-2r} (-1)^r 2^{-r} x^{-3r}$
$= \binom{10}{r} (-1)^r 2^{10-2r} x^{20-5r}$
For the term independent of $x$:
$20 – 5r = 0 \implies 5r = 20 \implies r = 4$
Therefore, the term independent of $x$ is the $(4+1) = 5\text{th}$ term.
Value of the term $= \binom{10}{4} (-1)^4 2^{10-2(4)} = 210 \times 1 \times 2^2 = 210 \times 4 = 840$
Answer: $840$
c) If the coefficient of $k^3$ in the expansion of $B$ is $30$, find the value of $x$.
Solution:
$B = \left(3k – \frac{2x}{3}\right)^5$
General term $T_{r+1} = \binom{5}{r} (3k)^{5-r} \left(-\frac{2x}{3}\right)^r$
To get the $k^3$ term from $(3k)^{5-r} \implies 5 – r = 3 \implies r = 2$
When $r = 2$, the term is:
$T_3 = \binom{5}{2} (3k)^3 \left(-\frac{2x}{3}\right)^2$
$= 10 \cdot 27k^3 \cdot \frac{4x^2}{9}$
$= 120 x^2 k^3$
Coefficient of $k^3 = 120 x^2$
According to the question,
$120 x^2 = 30$
$\implies x^2 = \frac{30}{120} = \frac{1}{4}$
$\implies x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$
Answer: $x = \pm \frac{1}{2}$
Short-Answer Questions:
a) Find the expansion of $(1 – 2x^2)^5$ using Pascal’s Triangle rule.
Solution:
Coefficients for $n = 5$: $1, 5, 10, 10, 5, 1$
$(1 – 2x^2)^5 = 1(1) + 5(1)(-2x^2) + 10(1)(-2x^2)^2 + 10(1)(-2x^2)^3 + 5(1)(-2x^2)^4 + 1(-2x^2)^5$
$= 1 – 10x^2 + 10(4x^4) – 10(8x^6) + 5(16x^8) – 32x^{10}$
$= 1 – 10x^2 + 40x^4 – 80x^6 + 80x^8 – 32x^{10}$
b) Determine coefficient of $x^3$ in the expansion of $\left(1 + \frac{x}{2}\right)^8$.
Solution:
Term containing $x^3$ is $T_4 = \binom{8}{3} (1)^{8-3} \left(\frac{x}{2}\right)^3 = 56 \times \frac{x^3}{8} = 7x^3$
Coefficient $= 7$
Answer: $7$
c) In the expansion of $\left(2k – \frac{x}{2}\right)^5$ coefficient of $k^3$ is $720$, find the value of $x$.
Solution:
To get the $k^3$ term from $(2k)^{5-r} \implies 5 – r = 3 \implies r = 2$
The term is $T_3 = \binom{5}{2} (2k)^3 \left(-\frac{x}{2}\right)^2 = 10 \cdot 8k^3 \cdot \frac{x^2}{4} = 20x^2 k^3$
Coefficient of $k^3 = 20x^2$
According to the question,
$20x^2 = 720 \implies x^2 = 36 \implies x = \pm 6$
Answer: $x = \pm 6$
d) If $\binom{n}{5} = \binom{n}{7}$, find the value of $n$.
Solution:
We know that if $\binom{n}{x} = \binom{n}{y}$, then $x = y$ or $n = x + y$.
Here $5 \neq 7$, therefore:
$n = 5 + 7 = 12$
Answer: $n = 12$