SL 1.9 Question Bank

Solution:

Step 1: Apply binomial theorem

\((x + 3)^4 = \sum_{r=0}^{4} \binom{4}{r} x^{4-r} 3^r\)

Step 2: Calculate each term

\(r = 0\): \(\binom{4}{0} x^4 3^0 = 1 \times x^4 \times 1 = x^4\)

\(r = 1\): \(\binom{4}{1} x^3 3^1 = 4 \times x^3 \times 3 = 12x^3\)

\(r = 2\): \(\binom{4}{2} x^2 3^2 = 6 \times x^2 \times 9 = 54x^2\)

\(r = 3\): \(\binom{4}{3} x^1 3^3 = 4 \times x \times 27 = 108x\)

\(r = 4\): \(\binom{4}{4} x^0 3^4 = 1 \times 1 \times 81 = 81\)

Step 3: Combine terms

\((x + 3)^4 = x^4 + 12x^3 + 54x^2 + 108x + 81\)

✅ Answer: \(x^4 + 12x^3 + 54x^2 + 108x + 81\)

Solution:

Step 1: Write general term

\(T_{r+1} = \binom{6}{r} (2x)^{6-r} (-1)^r\)

\(= \binom{6}{r} 2^{6-r} x^{6-r} (-1)^r\)

Step 2: Find r for \(x^3\) term

For \(x^3\): \(6-r = 3\), so \(r = 3\)

Step 3: Calculate coefficient

Coefficient = \(\binom{6}{3} \times 2^{6-3} \times (-1)^3\)

\(= 20 \times 2^3 \times (-1) = 20 \times 8 \times (-1) = -160\)

✅ Answer: -160

Solution:

Step 1: Determine middle term

For \((a + 2b)^8\), there are \(8 + 1 = 9\) terms

Since n = 8 is even, the middle term is the 5th term

Step 2: Find the 5th term

The 5th term corresponds to \(r = 4\)

\(T_5 = T_{4+1} = \binom{8}{4} a^{8-4} (2b)^4\)

Step 3: Calculate

\(T_5 = \binom{8}{4} a^4 (2b)^4 = 70 \times a^4 \times 16b^4 = 1120a^4b^4\)

✅ Answer: \(1120a^4b^4\)

Solution:

Step 1: Apply binomial theorem

\((1 – 2x)^3 = \sum_{r=0}^{3} \binom{3}{r} 1^{3-r} (-2x)^r\)

Step 2: Calculate each term

\(r = 0\): \(\binom{3}{0} (-2x)^0 = 1 \times 1 = 1\)

\(r = 1\): \(\binom{3}{1} (-2x)^1 = 3 \times (-2x) = -6x\)

\(r = 2\): \(\binom{3}{2} (-2x)^2 = 3 \times 4x^2 = 12x^2\)

\(r = 3\): \(\binom{3}{3} (-2x)^3 = 1 \times (-8x^3) = -8x^3\)

Step 3: Combine terms

\((1 – 2x)^3 = 1 – 6x + 12x^2 – 8x^3\)

✅ Answer: \(1 – 6x + 12x^2 – 8x^3\)

Solution:

Step 1: Identify the required term

The 4th term corresponds to \(r = 3\) (since \(T_{r+1}\) gives the (r+1)th term)

Step 2: Use general term formula

\(T_4 = T_{3+1} = \binom{7}{3} x^{7-3} (-2y)^3\)

Step 3: Calculate

\(T_4 = \binom{7}{3} x^4 (-2y)^3 = 35 \times x^4 \times (-8y^3)\)

\(= 35 \times (-8) \times x^4y^3 = -280x^4y^3\)

✅ Answer: \(-280x^4y^3\)

Solution:

Step 1: Write the binomial coefficient formula

\(\binom{n}{3} = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)}{3!} = \frac{n(n-1)(n-2)}{6}\)

Step 2: Set up equation

\(\frac{n(n-1)(n-2)}{6} = 35\)

Step 3: Solve

\(n(n-1)(n-2) = 210\)

Try n = 7: \(7 \times 6 \times 5 = 210\) ✓

Step 4: Verify

\(\binom{7}{3} = \frac{7 \times 6 \times 5}{6} = \frac{210}{6} = 35\) ✓

✅ Answer: n = 7

Solution:

Step 1: Express in binomial form

\((1.02)^8 = (1 + 0.02)^8\)

Step 2: Apply binomial theorem

\((1 + 0.02)^8 = \sum_{r=0}^{8} \binom{8}{r} (0.02)^r\)

Step 3: Calculate significant terms

\(r = 0\): \(\binom{8}{0} (0.02)^0 = 1\)

\(r = 1\): \(\binom{8}{1} (0.02)^1 = 8 \times 0.02 = 0.16\)

\(r = 2\): \(\binom{8}{2} (0.02)^2 = 28 \times 0.0004 = 0.0112\)

\(r = 3\): \(\binom{8}{3} (0.02)^3 = 56 \times 0.000008 = 0.000448\)

Step 4: Sum and round

\((1.02)^8 ≈ 1 + 0.16 + 0.0112 + 0.000448 + … ≈ 1.1717\)

✅ Answer: 1.1717

Solution:

Step 1: Write general term

\(T_{r+1} = \binom{12}{r} (x^2)^{12-r} \left(\frac{2}{x}\right)^r\)

Step 2: Simplify

\(T_{r+1} = \binom{12}{r} x^{2(12-r)} \frac{2^r}{x^r} = \binom{12}{r} 2^r x^{24-2r-r}\)

\(= \binom{12}{r} 2^r x^{24-3r}\)

Step 3: Find r for term independent of x

For independence: \(24 – 3r = 0\)

\(3r = 24\), so \(r = 8\)

Step 4: Calculate coefficient

Coefficient = \(\binom{12}{8} \times 2^8 = 495 \times 256 = 126720\)

✅ Answer: 126,720

Solution:

Step 1: Use substitution method

To find sum of coefficients, substitute x = 1 and y = 1

Step 2: Evaluate

Sum = \((2(1) – 3(1))^{10} = (2 – 3)^{10} = (-1)^{10} = 1\)

Step 3: Verify understanding

This means the positive and negative coefficients in the expansion almost cancel out,

leaving a sum of 1.

✅ Answer: 1

Solution:

Step 1: Identify the relevant term

General term: \(T_{r+1} = \binom{9}{r} x^{9-r} k^r\)

For \(x^4\): \(9-r = 4\), so \(r = 5\)

Step 2: Write coefficient equation

Coefficient of \(x^4\) = \(\binom{9}{5} k^5 = 3024\)

Step 3: Calculate binomial coefficient

\(\binom{9}{5} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4!} = \frac{3024}{24} = 126\)

Step 4: Solve for k

\(126k^5 = 3024\)

\(k^5 = \frac{3024}{126} = 24\)

\(k = \sqrt[5]{24} ≈ 1.89\) (positive root)

Step 5: Check for exact values

Since 24 is not a perfect fifth power, k = \(\sqrt[5]{24}\)

✅ Answer: \(k = \sqrt[5]{24}\)

Complete solution:

(a) First four terms:

\((1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \cdots\)

\(= 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3 + \cdots\)

(b) Finding n:

Coefficient of \(x^2\) = \(\binom{n}{2} = \frac{n(n-1)}{2} = 21\)

\(n(n-1) = 42\)

\(n^2 – n – 42 = 0\)

\((n-7)(n+6) = 0\)

Since n must be positive: n = 7

(c) Coefficient of \(x^5\) when n = 7:

Coefficient = \(\binom{7}{5} = \binom{7}{2} = 21\)

(d) Showing equality:

Using symmetry property: \(\binom{n}{r} = \binom{n}{n-r}\)

\(\binom{7}{5} = \binom{7}{7-5} = \binom{7}{2}\)

Therefore, coefficient of \(x^5\) = coefficient of \(x^2\) = 21

✅ Final Answers:
(a) \(1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3\)
(b) n = 7
(c) 21
(d) Proven by symmetry property

Complete solution:

(a) Probability distribution:

Using \((p + q)^n\) where p = 0.6 (prob of A), q = 0.4 (prob of a), n = 2

\((0.6 + 0.4)^2 = (0.6)^2 + 2(0.6)(0.4) + (0.4)^2\)

\(= 0.36 + 0.48 + 0.16\)

P(2 A’s) = 0.36, P(1 A) = 0.48, P(0 A’s) = 0.16

(b) Probability of exactly one A:

P(exactly 1 A) = 0.48 (genotype Aa)

(c) Expected number with genotype Aa:

Expected number = 5 × P(Aa) = 5 × 0.48 = 2.4

(d) Expansion and interpretation:

\((0.6 + 0.4)^{10} = 1^{10} = 1\)

Biological interpretation: This represents the total probability

across all possible outcomes when examining allele inheritance

in 10 independent events. The sum equals 1, confirming

that probabilities of all possible outcomes sum to certainty.

✅ Final Answers:
(a) P(AA) = 0.36, P(Aa) = 0.48, P(aa) = 0.16
(b) 0.48
(c) 2.4 offspring
(d) 1; represents total probability = certainty

Complete solution:

(a) Row 6:

1, 6, 15, 20, 15, 6, 1

(b) Sum of entries in row n:

Row n contains: \(\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, …, \binom{n}{n}\)

Sum = \(\sum_{r=0}^{n} \binom{n}{r}\)

Using binomial theorem with \((1+1)^n\):

\((1+1)^n = \sum_{r=0}^{n} \binom{n}{r} 1^{n-r} 1^r = \sum_{r=0}^{n} \binom{n}{r} = 2^n\)

(c) Alternating sum formula:

Alternating sum = \(\binom{n}{0} – \binom{n}{1} + \binom{n}{2} – \binom{n}{3} + …\)

\(= \sum_{r=0}^{n} (-1)^r \binom{n}{r}\)

Using binomial theorem with \((1-1)^n\):

\((1-1)^n = \sum_{r=0}^{n} \binom{n}{r} 1^{n-r} (-1)^r = \sum_{r=0}^{n} (-1)^r \binom{n}{r}\)

Therefore: alternating sum = \(0^n\)

(d) Verification for row 6:

Sum: 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = \(2^6\) ✓

Alternating: 1 – 6 + 15 – 20 + 15 – 6 + 1 = 0 = \(0^6\) ✓

(e) When n = 0:

Sum formula: \(2^0 = 1\) (correct, since row 0 has only entry 1)

Alternating formula: \(0^0 = 1\) (by convention)

Significance: Both give the same result (1) for the base case,

showing the consistency of the mathematical definitions.

✅ Final Answers:
(a) 1, 6, 15, 20, 15, 6, 1
(b) Sum = \(2^n\)
(c) Alternating sum = \(0^n\)
(d) 64 and 0 respectively
(e) Both equal 1, showing base case consistency

Complete solution:

(a) General probability expression:

Using binomial probability:

\(P(X = k) = \binom{n}{k} (0.8)^k (0.2)^{n-k}\)

(b) Probability of exactly 4 successes when n = 5:

\(P(X = 4) = \binom{5}{4} (0.8)^4 (0.2)^1\)

\(= 5 \times 0.4096 \times 0.2 = 5 \times 0.08192 = 0.4096\)

(c) Most likely number of successes:

Calculate all probabilities:

\(P(X = 0) = \binom{5}{0} (0.8)^0 (0.2)^5 = 1 \times 1 \times 0.00032 = 0.00032\)

\(P(X = 1) = \binom{5}{1} (0.8)^1 (0.2)^4 = 5 \times 0.8 \times 0.0016 = 0.0064\)

\(P(X = 2) = \binom{5}{2} (0.8)^2 (0.2)^3 = 10 \times 0.64 \times 0.008 = 0.0512\)

\(P(X = 3) = \binom{5}{3} (0.8)^3 (0.2)^2 = 10 \times 0.512 \times 0.04 = 0.2048\)

\(P(X = 4) = 0.4096\) (from part b)

\(P(X = 5) = \binom{5}{5} (0.8)^5 (0.2)^0 = 1 \times 0.32768 \times 1 = 0.32768\)

Maximum probability is 0.4096, so most likely: 4 successes

(d) At least 4 successes:

\(P(X \geq 4) = P(X = 4) + P(X = 5)\)

\(= 0.4096 + 0.32768 = 0.73728\)

Comment: 73.7% probability suggests good viability,

though there’s still a 26.3% chance of insufficient success.

✅ Final Answers:
(a) \(P(X = k) = \binom{n}{k} (0.8)^k (0.2)^{n-k}\)
(b) 0.4096
(c) 4 successes (highest probability)
(d) 0.7373 or 73.73% – reasonably viable