GSEB Class 11 Maths Solutions Chapter 8 Binomial Theorem Exercise 8.1

Expand Each of the Expressions in Questions 1 – 5:

Question 1. \( (1 - 2x)^5 \)
Answer: We can expand \( (1 - 2x)^5 \) using the binomial theorem. The formula for binomial expansion is \( (a+b)^n = \sum_{k=0}^{n} {}^nC_k a^{n-k} b^k \). Here, \( a = 1 \), \( b = -2x \), and \( n = 5 \).
\( (1 - 2x)^5 = {}^5C_0(1)^5(-2x)^0 + {}^5C_1(1)^4(-2x)^1 + {}^5C_2(1)^3(-2x)^2 + {}^5C_3(1)^2(-2x)^3 + {}^5C_4(1)^1(-2x)^4 + {}^5C_5(1)^0(-2x)^5 \)
\( = 1 \cdot 1 \cdot 1 + 5 \cdot 1 \cdot (-2x) + \frac{5 \cdot 4}{1 \cdot 2} \cdot 1 \cdot (4x^2) + \frac{5 \cdot 4}{1 \cdot 2} \cdot 1 \cdot (-8x^3) + \frac{5}{1} \cdot 1 \cdot (16x^4) + 1 \cdot 1 \cdot (-32x^5) \)
\( = 1 - 10x + 10(4x^2) - 10(8x^3) + 5(16x^4) - 32x^5 \)
\( = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 \)
In simple words: To expand this expression, we use the binomial theorem. This involves finding the combinations of terms and their powers, then multiplying them out. We combine all the parts to get the final expanded form.

Exam Tip: Remember the signs of \( b \) in the binomial expansion. If \( b \) is negative, its odd powers will be negative, and even powers will be positive. Be careful with calculations of combinations and powers.

Question 2. \( \left(\frac{2}{x} - \frac{x}{2}\right)^5 \)
Answer: We expand \( \left(\frac{2}{x} - \frac{x}{2}\right)^5 \) using the binomial theorem. Here, \( a = \frac{2}{x} \), \( b = -\frac{x}{2} \), and \( n = 5 \).
\( \left(\frac{2}{x} - \frac{x}{2}\right)^5 = {}^5C_0\left(\frac{2}{x}\right)^5\left(-\frac{x}{2}\right)^0 + {}^5C_1\left(\frac{2}{x}\right)^4\left(-\frac{x}{2}\right)^1 + {}^5C_2\left(\frac{2}{x}\right)^3\left(-\frac{x}{2}\right)^2 + {}^5C_3\left(\frac{2}{x}\right)^2\left(-\frac{x}{2}\right)^3 + {}^5C_4\left(\frac{2}{x}\right)^1\left(-\frac{x}{2}\right)^4 + {}^5C_5\left(-\frac{x}{2}\right)^5 \)
\( = 1 \cdot \frac{32}{x^5} \cdot 1 + 5 \cdot \frac{16}{x^4} \cdot \left(-\frac{x}{2}\right) + 10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4} + 10 \cdot \frac{4}{x^2} \cdot \left(-\frac{x^3}{8}\right) + 5 \cdot \frac{2}{x} \cdot \frac{x^4}{16} + 1 \cdot \left(-\frac{x^5}{32}\right) \)
\( = \frac{32}{x^5} - \frac{80x}{x^4} + \frac{80x^2}{4x^3} - \frac{40x^3}{8x^2} + \frac{10x^4}{16x} - \frac{x^5}{32} \)
\( = 32x^{-5} - 40x^{-3} + 20x^{-1} - 5x + \frac{5}{8}x^3 - \frac{x^5}{32} \)
In simple words: This expansion follows the binomial theorem. We carefully apply the powers to both the numerator and denominator of the fractions and then simplify by combining terms and reducing fractions. Keep an eye on negative signs and powers of x.

Exam Tip: When dealing with fractional terms like \( \frac{1}{x} \) or \( x^{-1} \), remember to apply the exponents correctly to both the numerator and the denominator. Simplify all terms completely to get the final answer.

Question 3. \( (2x - 3)^6 \)
Answer: We expand \( (2x - 3)^6 \) using the binomial theorem. Here, \( a = 2x \), \( b = -3 \), and \( n = 6 \).
\( (2x - 3)^6 = {}^6C_0(2x)^6(-3)^0 + {}^6C_1(2x)^5(-3)^1 + {}^6C_2(2x)^4(-3)^2 + {}^6C_3(2x)^3(-3)^3 + {}^6C_4(2x)^2(-3)^4 + {}^6C_5(2x)^1(-3)^5 + {}^6C_6(2x)^0(-3)^6 \)
\( = 1 \cdot (64x^6) \cdot 1 + 6 \cdot (32x^5) \cdot (-3) + 15 \cdot (16x^4) \cdot 9 + 20 \cdot (8x^3) \cdot (-27) + 15 \cdot (4x^2) \cdot (81) + 6 \cdot (2x) \cdot (-243) + 1 \cdot 1 \cdot (729) \)
\( = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729 \)
In simple words: For this expansion, we follow the binomial formula. Each term gets a coefficient from Pascal's triangle, and the powers of \(2x\) go down while the powers of \(-3\) go up. We multiply everything and then combine the terms.

Exam Tip: Be especially careful with the negative sign when \( b \) is negative. Odd powers of a negative number result in negative values, which affects the sign of the overall term.

Question 4. \( \left(\frac{x}{3} + \frac{1}{x}\right)^5 \)
Answer: We expand \( \left(\frac{x}{3} + \frac{1}{x}\right)^5 \) using the binomial theorem. Here, \( a = \frac{x}{3} \), \( b = \frac{1}{x} \), and \( n = 5 \).
\( \left(\frac{x}{3} + \frac{1}{x}\right)^5 = {}^5C_0\left(\frac{x}{3}\right)^5\left(\frac{1}{x}\right)^0 + {}^5C_1\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right)^1 + {}^5C_2\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 + {}^5C_3\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 + {}^5C_4\left(\frac{x}{3}\right)^1\left(\frac{1}{x}\right)^4 + {}^5C_5\left(\frac{1}{x}\right)^5 \)
\( = 1 \cdot \frac{x^5}{243} \cdot 1 + 5 \cdot \frac{x^4}{81} \cdot \frac{1}{x} + 10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2} + 10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3} + 5 \cdot \frac{x}{3} \cdot \frac{1}{x^4} + 1 \cdot \frac{1}{x^5} \)
\( = \frac{x^5}{243} + \frac{5x^4}{81x} + \frac{10x^3}{27x^2} + \frac{10x^2}{9x^3} + \frac{5x}{3x^4} + \frac{1}{x^5} \)
\( = \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5} \)
In simple words: This problem asks us to expand an expression with fractions using the binomial theorem. We apply the powers to both parts of the fractions and then simplify by combining terms. Make sure to correctly handle the exponents of x in the numerator and denominator.

Exam Tip: When simplifying terms with different powers of \( x \) in the numerator and denominator, remember the rule \( \frac{x^m}{x^n} = x^{m-n} \). This will help you reduce the terms effectively.

Question 5. \( \left(x + \frac{1}{x}\right)^6 \)
Answer: We expand \( \left(x + \frac{1}{x}\right)^6 \) using the binomial theorem. Here, \( a = x \), \( b = \frac{1}{x} \), and \( n = 6 \).
\( \left(x + \frac{1}{x}\right)^6 = {}^6C_0(x)^6\left(\frac{1}{x}\right)^0 + {}^6C_1(x)^5\left(\frac{1}{x}\right)^1 + {}^6C_2(x)^4\left(\frac{1}{x}\right)^2 + {}^6C_3(x)^3\left(\frac{1}{x}\right)^3 + {}^6C_4(x)^2\left(\frac{1}{x}\right)^4 + {}^6C_5(x)^1\left(\frac{1}{x}\right)^5 + {}^6C_6(x)^0\left(\frac{1}{x}\right)^6 \)
\( = 1 \cdot x^6 \cdot 1 + 6 \cdot x^5 \cdot \frac{1}{x} + 15 \cdot x^4 \cdot \frac{1}{x^2} + 20 \cdot x^3 \cdot \frac{1}{x^3} + 15 \cdot x^2 \cdot \frac{1}{x^4} + 6 \cdot x \cdot \frac{1}{x^5} + 1 \cdot 1 \cdot \frac{1}{x^6} \)
\( = x^6 + 6x^4 + 15x^2 + 20 + 15x^{-2} + 6x^{-4} + x^{-6} \)
In simple words: To expand this expression, we use the binomial theorem. We carefully apply the powers to both \(x\) and \(1/x\), remembering that when \(x\) and \(1/x\) are multiplied, their exponents subtract. This process simplifies to terms with decreasing powers of \(x\).

Exam Tip: Notice the symmetry in the coefficients of the binomial expansion. For \( (a+b)^n \), the coefficients for \( a^{n-k} b^k \) and \( a^k b^{n-k} \) are the same. This can help you verify your calculations.

Using Binomial Theorem, Evaluate Each of the Following Questions 6 – 9:

Question 6. \( (96)^3 \)
Answer: We can evaluate \( (96)^3 \) by writing it as \( (100 - 4)^3 \) and applying the binomial theorem. Here, \( a = 100 \), \( b = -4 \), and \( n = 3 \).
\( (100 - 4)^3 = {}^3C_0(100)^3(-4)^0 + {}^3C_1(100)^2(-4)^1 + {}^3C_2(100)^1(-4)^2 + {}^3C_3(100)^0(-4)^3 \)
\( = 1 \cdot (1000000) \cdot 1 + 3 \cdot (10000) \cdot (-4) + 3 \cdot (100) \cdot (16) + 1 \cdot 1 \cdot (-64) \)
\( = 1000000 - 120000 + 4800 - 64 \)
\( = 880000 + 4800 - 64 \)
\( = 884800 - 64 \)
\( = 884736 \)
In simple words: To calculate \(96^3\), we rewrite it as \((100-4)^3\). Then we use the binomial theorem, expanding each part, multiplying them, and adding/subtracting to get the final answer. This way, large numbers become easier to handle.

Exam Tip: Expressing numbers close to powers of 10 (like 96 as 100-4 or 102 as 100+2) simplifies binomial expansion because powers of 10 are easy to compute.

Question 7. \( (102)^5 \)
Answer: We can evaluate \( (102)^5 \) by writing it as \( (100 + 2)^5 \) and applying the binomial theorem. Here, \( a = 100 \), \( b = 2 \), and \( n = 5 \).
\( (100 + 2)^5 = {}^5C_0(100)^5(2)^0 + {}^5C_1(100)^4(2)^1 + {}^5C_2(100)^3(2)^2 + {}^5C_3(100)^2(2)^3 + {}^5C_4(100)^1(2)^4 + {}^5C_5(100)^0(2)^5 \)
\( = 1 \cdot (10000000000) \cdot 1 + 5 \cdot (100000000) \cdot 2 + 10 \cdot (1000000) \cdot 4 + 10 \cdot (10000) \cdot 8 + 5 \cdot (100) \cdot 16 + 1 \cdot 1 \cdot 32 \)
\( = 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 \)
\( = 11040808032 \)
In simple words: To find \(102^5\), we change it to \((100+2)^5\). Using the binomial theorem, we expand each term, multiply the parts together, and then add them up to get the final numerical result.

Exam Tip: Clearly write out each term of the expansion before calculating. This helps to prevent errors in powers, combinations, and multiplication, especially with larger numbers.

Question 8. \( (101)^4 \)
Answer: We can evaluate \( (101)^4 \) by writing it as \( (100 + 1)^4 \) and applying the binomial theorem. Here, \( a = 100 \), \( b = 1 \), and \( n = 4 \).
\( (100 + 1)^4 = {}^4C_0(100)^4(1)^0 + {}^4C_1(100)^3(1)^1 + {}^4C_2(100)^2(1)^2 + {}^4C_3(100)^1(1)^3 + {}^4C_4(100)^0(1)^4 \)
\( = 1 \cdot (100000000) \cdot 1 + 4 \cdot (1000000) \cdot 1 + 6 \cdot (10000) \cdot 1 + 4 \cdot (100) \cdot 1 + 1 \cdot 1 \cdot 1 \)
\( = 100000000 + 4000000 + 60000 + 400 + 1 \)
\( = 104060401 \)
In simple words: To calculate \(101^4\), we treat it as \((100+1)^4\). We then use the binomial theorem to expand the expression. After finding each term by multiplying coefficients and powers, we add them all up to find the final numerical value.

Exam Tip: When \( b=1 \), the powers of \( b \) simplify to 1, making calculations easier. This is a common strategy to simplify expressions for binomial expansion.

Question 9. \( (99)^5 \)
Answer: We can evaluate \( (99)^5 \) by writing it as \( (100 - 1)^5 \) and applying the binomial theorem. Here, \( a = 100 \), \( b = -1 \), and \( n = 5 \).
\( (100 - 1)^5 = {}^5C_0(100)^5(-1)^0 + {}^5C_1(100)^4(-1)^1 + {}^5C_2(100)^3(-1)^2 + {}^5C_3(100)^2(-1)^3 + {}^5C_4(100)^1(-1)^4 + {}^5C_5(100)^0(-1)^5 \)
\( = 1 \cdot (10000000000) \cdot 1 + 5 \cdot (100000000) \cdot (-1) + 10 \cdot (1000000) \cdot 1 + 10 \cdot (10000) \cdot (-1) + 5 \cdot (100) \cdot 1 + 1 \cdot 1 \cdot (-1) \)
\( = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1 \)
\( = 9509900499 \)
In simple words: To calculate \(99^5\), we express it as \((100-1)^5\). We then use the binomial theorem, expanding each term and carefully managing the signs due to the negative 1. Finally, we sum all the terms to get the result.

Exam Tip: When \( b=-1 \), the terms in the expansion will alternate in sign, starting with positive. This is a crucial detail to remember to avoid errors in calculation.

Question 10. Using Binomial Theorem, indicate which number is larger \( (1.1)^{10000} \) or \( 1000 \)?
Answer: We need to compare \( (1.1)^{10000} \) and \( 1000 \). Let's use the binomial theorem to expand \( (1.1)^{10000} \):
\( (1.1)^{10000} = (1 + 0.1)^{10000} \)
Using the binomial expansion formula \( (a+b)^n = {}^nC_0 a^n b^0 + {}^nC_1 a^{n-1} b^1 + {}^nC_2 a^{n-2} b^2 + \dots \)
Here, \( a = 1 \), \( b = 0.1 \), and \( n = 10000 \).
\( (1 + 0.1)^{10000} = {}^{10000}C_0 (1)^{10000} (0.1)^0 + {}^{10000}C_1 (1)^{9999} (0.1)^1 + \text{other positive terms} \)
\( = 1 \cdot 1 \cdot 1 + 10000 \cdot 1 \cdot 0.1 + \text{other positive terms} \)
\( = 1 + 1000 + \text{other positive terms} \)
\( = 1001 + \text{other positive terms} \)
Since all the remaining terms in the binomial expansion are positive, \( (1.1)^{10000} \) is definitely greater than \( 1001 \).
Therefore, \( (1.1)^{10000} > 1000 \).
In simple words: To find out which number is bigger, we expand \((1.1)^{10000}\) using the binomial theorem. We split it into \((1+0.1)^{10000}\). The first two parts of the expansion give \(1 + 1000\), which is \(1001\). Since all other parts are positive, the total value will be more than \(1001\). So, \((1.1)^{10000}\) is larger than \(1000\).

Exam Tip: For problems comparing magnitudes of numbers, especially with large exponents, binomial expansion is a powerful tool. You often only need to compute the first few terms to draw a conclusion, particularly when comparing to a smaller number.

Question 11. Find \( (a + b)^4 – (a – b)^4 \). Hence, evaluate \( (\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} – \sqrt{2})^4 \).
Answer:First, let's find the expansion of \( (a+b)^4 \) and \( (a-b)^4 \). Using the binomial theorem:
\( (a + b)^4 = {}^4C_0 a^4 b^0 + {}^4C_1 a^3 b^1 + {}^4C_2 a^2 b^2 + {}^4C_3 a^1 b^3 + {}^4C_4 a^0 b^4 \)
\( = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \) (Equation 1)
\( (a - b)^4 = {}^4C_0 a^4 (-b)^0 + {}^4C_1 a^3 (-b)^1 + {}^4C_2 a^2 (-b)^2 + {}^4C_3 a^1 (-b)^3 + {}^4C_4 a^0 (-b)^4 \)
\( = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 \) (Equation 2)
Now, we subtract Equation 2 from Equation 1:
\( (a + b)^4 - (a - b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4) \)
\( = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 - a^4 + 4a^3b - 6a^2b^2 + 4ab^3 - b^4 \)
\( = 8a^3b + 8ab^3 \)
\( = 8ab(a^2 + b^2) \)
Next, we evaluate \( (\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} – \sqrt{2})^4 \) by substituting \( a = \sqrt{3} \) and \( b = \sqrt{2} \) into the derived expression \( 8ab(a^2 + b^2) \):
\( 8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2) \)
\( = 8\sqrt{6}(3 + 2) \)
\( = 8\sqrt{6}(5) \)
\( = 40\sqrt{6} \)
In simple words: First, we expand \((a+b)^4\) and \((a-b)^4\) using the binomial theorem. Then, we subtract the second expansion from the first one. This simplifies to \(8ab(a^2+b^2)\). Finally, we replace \(a\) with \(\sqrt{3}\) and \(b\) with \(\sqrt{2}\) in this simplified expression to get the numerical answer.

Exam Tip: When asked to "hence evaluate," it means you should use the result from the first part of the question. Do not re-expand the specific numerical values from scratch. Substituting into the general algebraic result is faster and less prone to errors.

Question 12. Find \( (x + 1)^6 + (x – 1)^6 \). Hence, or otherwise, evaluate \( (\sqrt{2} + 1)^6 + (\sqrt{2} – 1)^6 \).
Answer:First, let's find the expansion of \( (x+1)^6 \) and \( (x-1)^6 \). Using the binomial theorem:
\( (x + 1)^6 = {}^6C_0 x^6 (1)^0 + {}^6C_1 x^5 (1)^1 + {}^6C_2 x^4 (1)^2 + {}^6C_3 x^3 (1)^3 + {}^6C_4 x^2 (1)^4 + {}^6C_5 x^1 (1)^5 + {}^6C_6 x^0 (1)^6 \)
\( = x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1 \) (Equation 1)
\( (x - 1)^6 = {}^6C_0 x^6 (-1)^0 + {}^6C_1 x^5 (-1)^1 + {}^6C_2 x^4 (-1)^2 + {}^6C_3 x^3 (-1)^3 + {}^6C_4 x^2 (-1)^4 + {}^6C_5 x^1 (-1)^5 + {}^6C_6 x^0 (-1)^6 \)
\( = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 \) (Equation 2)
Now, we add Equation 1 and Equation 2:
\( (x + 1)^6 + (x - 1)^6 = (x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1) + (x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1) \)
\( = 2x^6 + 30x^4 + 30x^2 + 2 \)
\( = 2(x^6 + 15x^4 + 15x^2 + 1) \)
Next, we evaluate \( (\sqrt{2} + 1)^6 + (\sqrt{2} – 1)^6 \) by substituting \( x = \sqrt{2} \) into the derived expression \( 2(x^6 + 15x^4 + 15x^2 + 1) \):
\( = 2((\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1) \)
\( = 2(2^3 + 15 \cdot 2^2 + 15 \cdot 2 + 1) \)
\( = 2(8 + 15 \cdot 4 + 30 + 1) \)
\( = 2(8 + 60 + 30 + 1) \)
\( = 2(99) \)
\( = 198 \)
In simple words: First, we expand \((x+1)^6\) and \((x-1)^6\) using the binomial theorem. Adding these two expansions cancels out some terms and simplifies to \(2(x^6 + 15x^4 + 15x^2 + 1)\). Then, we substitute \(x = \sqrt{2}\) into this simplified expression. We calculate the powers of \(\sqrt{2}\) and perform the multiplications and additions to get the final numerical result.

Exam Tip: When adding or subtracting binomial expansions where one term is negative, remember that terms with odd powers of the negative term will cancel out when adding, and terms with even powers will cancel out when subtracting.

Question 13. Show that \( 9^{n+1} - 8n - 9 \) is divisible by 64, whenever \( n \) is a positive integer.
Answer: We need to show that \( 9^{n+1} - 8n - 9 \) is divisible by 64. Let's consider the binomial expansion of \( (1 + x)^{n+1} \):
\( (1 + x)^{n+1} = {}^{n+1}C_0 (1)^{n+1} (x)^0 + {}^{n+1}C_1 (1)^{n} (x)^1 + {}^{n+1}C_2 (1)^{n-1} (x)^2 + {}^{n+1}C_3 (1)^{n-2} (x)^3 + \dots + {}^{n+1}C_{n+1} (1)^0 (x)^{n+1} \)
We know that \( {}^{n+1}C_0 = 1 \) and \( {}^{n+1}C_1 = n+1 \). Substituting \( x = 8 \) into the expansion:
\( (1 + 8)^{n+1} = {}^{n+1}C_0 (1)^{n+1} (8)^0 + {}^{n+1}C_1 (1)^{n} (8)^1 + {}^{n+1}C_2 (1)^{n-1} (8)^2 + {}^{n+1}C_3 (1)^{n-2} (8)^3 + \dots + {}^{n+1}C_{n+1} (1)^0 (8)^{n+1} \)
\( 9^{n+1} = 1 \cdot 1 \cdot 1 + (n+1) \cdot 1 \cdot 8 + {}^{n+1}C_2 \cdot 1 \cdot 8^2 + {}^{n+1}C_3 \cdot 1 \cdot 8^3 + \dots + {}^{n+1}C_{n+1} \cdot 1 \cdot 8^{n+1} \)
\( 9^{n+1} = 1 + 8(n+1) + {}^{n+1}C_2 \cdot 64 + {}^{n+1}C_3 \cdot 8^3 + \dots + {}^{n+1}C_{n+1} \cdot 8^{n+1} \)
\( 9^{n+1} = 1 + 8n + 8 + {}^{n+1}C_2 \cdot 64 + {}^{n+1}C_3 \cdot 8^3 + \dots + {}^{n+1}C_{n+1} \cdot 8^{n+1} \)
Now, rearrange the terms to isolate \( 9^{n+1} - 8n - 9 \):
\( 9^{n+1} - 8n - 9 = {}^{n+1}C_2 \cdot 64 + {}^{n+1}C_3 \cdot 8^3 + \dots + {}^{n+1}C_{n+1} \cdot 8^{n+1} \)
Notice that every term on the right-hand side has a factor of \( 64 \) (since \( 8^2 = 64 \), and \( 8^3 = 8 \cdot 64 \), and so on).
So, \( 9^{n+1} - 8n - 9 = 64 [{}^{n+1}C_2 + {}^{n+1}C_3 \cdot 8 + \dots + {}^{n+1}C_{n+1} \cdot 8^{n-1}] \)
Since the expression in the square brackets is an integer (sum of products of integers), we can conclude that \( 9^{n+1} - 8n - 9 \) is a multiple of 64.
Therefore, \( 9^{n+1} - 8n - 9 \) is divisible by 64 for any positive integer \( n \). (Proved)
In simple words: To prove that \(9^{n+1} - 8n - 9\) can be divided by \(64\), we use the binomial expansion of \((1+x)^{n+1}\). We substitute \(x=8\) into the expansion. After rearranging the terms, we find that the expression is equal to \(64\) multiplied by a sum of integers. This means the original expression is always a multiple of \(64\), proving it is divisible by \(64\).

Exam Tip: When proving divisibility using binomial expansion, identify a suitable expression (often \( (1+k)^n \)) and choose a value for \( k \) that relates to the divisor. Then, expand and isolate the terms to show the desired factor.

Question 14. Prove that \( \sum_{r=0}^{n} 3^r {}^n C_r = 4^n \).
Answer: We need to prove the identity \( \sum_{r=0}^{n} 3^r {}^n C_r = 4^n \). We know the binomial theorem for the expansion of \( (a+b)^n \):
\( (a+b)^n = \sum_{r=0}^{n} {}^nC_r a^{n-r} b^r \)
Let's consider the expansion of \( (1+3)^n \). Here, we have \( a = 1 \) and \( b = 3 \).
Substituting these values into the binomial theorem formula:
\( (1+3)^n = \sum_{r=0}^{n} {}^nC_r (1)^{n-r} (3)^r \)
Since \( (1)^{n-r} = 1 \) for any integer \( n-r \), the expression simplifies to:
\( (1+3)^n = \sum_{r=0}^{n} {}^nC_r 3^r \)
Calculating the left-hand side:
\( (1+3)^n = (4)^n = 4^n \)
So, we have:
\( 4^n = \sum_{r=0}^{n} {}^nC_r 3^r \)
Therefore, \( \sum_{r=0}^{n} 3^r {}^n C_r = 4^n \). (Proved)
In simple words: To prove this identity, we start with the binomial theorem, which explains how \((a+b)^n\) expands. We then substitute \(a=1\) and \(b=3\) into the formula. This makes the left side \((1+3)^n\), which simplifies to \(4^n\). The right side becomes the sum we wanted to prove, showing that both sides are equal.

Exam Tip: This type of problem often tests your understanding of the general form of the binomial theorem. Recognizing the values of \( a \) and \( b \) that transform the theorem into the given sum is key to a quick solution.