1. To prove 1 + 3 + 5 + ... + (2n - 1) = n2 by induction, which line correctly uses the k-case to prove the (k + 1)-case?
Assume the sum of the first k odd numbers is k2. The next odd number is 2k + 1, so the next sum is k2 + 2k + 1 = (k + 1)2.
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Assume the sum of the first k odd numbers is k2. The next odd number is 2k + 1, so the next sum is k2 + 2k + 1 = (k + 1)2.
Since b . a = 11 and a . a = 10, proj_a b = (11/10)a. Hence p = b - proj_a b = (-13/10, 39/10), with magnitude √(1690)/10 = 13√10/10.
The x4 term is C(7, 4)(-2x)4 = 35 x 16x4 = 560x4, so the coefficient is 560.
Factor the equation as (2cos x - 1)(cos x + 1) = 0. Thus cos x = 1/2 or cos x = -1. On 0 ≤ x < 2π, these occur at π/3, 5π/3 and π, giving three solutions.
On 0 ≤ x ≤ 1, x is above x2. The area is the integral from 0 to 1 of x - x2, which is 1/2 - 1/3 = 1/6.
For a binomial random variable, E(X) = np. Here E(X) = 8 x 0.35 = 2.8, meaning the long-run average number of successes is 2.8 per eight trials.
HSC Year 12 Mathematics Extension 1 practice should give strong students enough structure to revise induction, vectors, binomial theorem, trigonometry, calculus, and probability. These examples preview that style before the no-login Extension 1 demo.
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