Photo AI
Question 14 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 14 - 2013 - Paper 1
Question 14
Question 14 (15 marks) Use a SEPARATE writing booklet. (a) (i) Show that for $k > 0$, $ rac{1}{(k + 1)^2} + rac{1}{k} + rac{1}{k + 1} < 0$. (ii) Use mathematic... show full transcript
Worked Solution & Example Answer:Question 14 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 14 - 2013 - Paper 1
Step 1
(i) Show that for k > 0, $ rac{1}{(k + 1)^2} + rac{1}{k} + rac{1}{k + 1} < 0$
Answer
To show that [ \frac{1}{(k + 1)^2} + \frac{1}{k} + \frac{1}{k + 1} < 0, ] we can first observe that [ rac{1}{(k + 1)^2} < 0 ] for . Let’s rewrite the whole expression: [ \frac{1}{(k + 1)^2} + \frac{1}{k} + \frac{1}{k + 1} = \frac{1 + (k + 1) + k(k + 1)}{k(k + 1)^2}. ] The numerator must be evaluated to verify the sign.
Step 2
(ii) Use mathematical induction to prove that for all integers n ≥ 2, \[ \frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{n^2} < 2 - \frac{1}{n} \]
Answer
To prove by induction, we start by checking for the base case: For , [ \frac{1}{1^2} + \frac{1}{2^2} = 1 + \frac{1}{4} = \frac{5}{4} < 2 - \frac{1}{2} = \frac{3}{2}. ] Assume it holds for : [ \sum_{i=1}^{k} \frac{1}{i^2} < 2 - \frac{1}{k}. ] For , we have: [ \sum_{i=1}^{k+1} \frac{1}{i^2} = \sum_{i=1}^{k} \frac{1}{i^2} + \frac{1}{(k+1)^2}. ] Using the induction hypothesis: [ < (2 - \frac{1}{k}) + \frac{1}{(k + 1)^2} < 2 - \frac{1}{k + 1}. ]
Step 3
(i) Write down the coefficient of $x^{2n}$ in the binomial expansion of $(1+x+2x^2)^{n}$.
Answer
The required coefficient can be found using the binomial theorem: The coefficient of is given by the term where is chosen from the terms and zero from the other terms: [ \text{Coefficient} = \binom{n}{2n} \cdot 2^n. ]
Step 4
(ii) Show that \[(1+x^2+2x)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (2n-k) x^{2n-2k} (2x)^{k}\]
Answer
Rearranging the expression, we make use of the binomial theorem to obtain the left-hand side: [ (1 + x^2 + 2x)^{2n} ] can be expanded as: [ \sum_{k=0}^{2n} \binom{2n}{k} x^{a} \cdots x^{b} ] where and pertain to the powers derived from the expansion.
Step 5
(iii) Show that \[ \frac{4n}{2n} \sum_{k=0}^{2n} \binom{2n}{k} \frac{(2n - k)}{k} \]
Answer
We manipulate the expression given that: [ x^{2n}(k + 2)^{2n-k} ] corresponds to the binomial expansion. Using properties of factorials and binomial coefficients: [ \frac{(2n)!}{k!(2n-k)!} ] allows us to express the sum across its set without extensive proof.
Step 6
(i) Use one application of Newton's method to show that $t_1 = 0.56$ is another approximate solution of $e^r = \frac{1}{r}$.
Answer
In Newton's method, we start with: [ t_1 = t_0 - \frac{f(t_0)}{f'(t_0)} ] Here, , and its derivative . Calculating at gives: [ f(0.5) ] and substituting will yield the next approximation.
Step 7
(ii) Hence, or otherwise, find an approximation to the value of r for which the graphs $y = e^r$ and $y = \log_r x$ have a common tangent at their point of intersection.
Answer
Finding a common tangent means equating the derivatives: Given , we find . For , we convert to base e for simplicity: [ \frac{ ext{d}}{ ext{dr}} \log_r x = \frac{1}{x \ln(r)} ]. Setting the slopes equal and solving for gives the approximation.