ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: davidss on November 07, 2016, 10:05:13 pm

Title: Induction Help!
Post by: davidss on November 07, 2016, 10:05:13 pm: Hi, can someone help with 14, 24 and 31 please. Thanks in advanced :D
Title: Re: Induction Help!
Post by: RuiAce on November 07, 2016, 10:11:10 pm: $\text{I'm only going to do Q31 for now as it's a bit weirder.}\\ \text{Post your own attempts at the rest and we can offer tips.}$
$\text{Note that some people will prefer expanding out the sigma into:}\\ \ln \frac21 +\ln \frac32 + \ln \frac43 + \dots + \ln \frac{n}{n-1}\\ \text{But here I won't.}\\ \text{When }n=2\text{ (statement not defined for }n=1)\\ LHS = \sum_{k=2}^2 \ln \frac{k}{k-1} = \ln 2 = RHS\\ \text{Hence true when }n=2$
$\text{Assuming true for }n=j\text{ gives the result}\\ \sum_{k=2}^j \ln \left(\frac{k}{k-1}\right)=\ln j\\ \text{RTP: The statement holds for }n=j+1\\ \text{i.e. }\sum_{k=2}^{j+1} \ln \left(\frac{k}{k-1}\right) = \ln (j+1)\\ \text{We will utilise the log law }\log A + \log B = \log(AB)$
$\begin{align*}LHS&=\sum_{k=2}^{j+1} \ln \left(\frac{k}{k-1}\right)\\ &=\sum_{k=2}^j \ln \left(\frac{k}{k-1}\right) + \ln \left(\frac{j+1}{j+1-1}\right)\\ &= \ln j + \ln \left(\frac{j+1}{j}\right) \tag{using assumption}\\ &= \ln \left(\frac{j+1}{j} \times j\right)\\ &= \ln (j+1)\\ &= RHS\end{align*}$
Title: Re: Induction Help!
Post by: Dolphax on November 08, 2016, 05:38:23 pm: $\text{Q24:}\\ \\ \sum_{r=1}^{n}(2r-1^3)=n^2(2n^2-1)\\ \\ ie. (2(1)-1)^3 + (2(2)-1)^3 + (2(3)-1)^3 + ... + (2n-1)^3 = n^2(2n^2-1)\\ \\ \text{Step 1: For n = 1}\\ \\ LHS = 1, RHS = 1 \text{(show your working here)} \\ \\ \text{Step 2: Assume true for n = k}\\ \\ (2(1)-1)^3 + (2(2)-1)^3 + (2(3)-1)^3 + ... + (2k-1)^3 = k^2(2k^2-1)\\ \\ \text{Step 3: Consider n = k + 1}\\ \\ RTP:(2(1)-1)^3 + (2(2)-1)^3 + (2(3)-1)^3 + ... + (2k-1)^3 + (2(k+1)-1)^3 = (k+1)^2(2(k+1)^2-1)\\ = (k+1)^2(2k^2+4k+2-1)\\ = (k^2+2k+1)(2k^2+4k+1)\\ = 2k^4+8k^3+11k^2+6k+1 \\ \\ \text{Now, LHS = } 2(1)-1)^3 + (2(2)-1)^3 + (2(3)-1)^3 + ... + (2k-1)^3 + (2(k+1)-1)^3\\ = k^2(2k^2-1) + (2(k+1)-1)^3 \text{ (using assumption)}\\ = k^2(2k^2-1)+(2k+1)^3\\ = 2k^4-k^2+(2k+1)(4k^2+4k+1)\\ = 2k^4-k^2+8k^3+8k^2+2k+4k^2+4k+1\\ = 2k^4+8k^3+11k^2+6k+1\\ = RHS\\$
Title: Re: Induction Help!
Post by: jamonwindeyer on November 08, 2016, 08:37:48 pm: Quote from: Dolphax on November 08, 2016, 05:38:23 pm
$\text{Q24:}\\ \\ \sum_{r=1}^{n}(2r-1^3)=n^2(2n^2-1)\\ \\ ie. (2(1)-1)^3 + (2(2)-1)^3 + (2(3)-1)^3 + ... + (2n-1)^3 = n^2(2n^2-1)\\ \\ \text{Step 1: For n = 1}\\ \\ LHS = 1, RHS = 1 \text{(show your working here)} ...$

Welcome to the forums Dolphax! ;D thanks for the awesome answer :)
Title: Re: Induction Help!
Post by: Mahan on November 19, 2016, 11:38:10 pm: Quote from: davidss on November 07, 2016, 10:05:13 pm
Hi, can someone help with 14, 24 and 31 please. Thanks in advanced :D

Q14:
Base case: n=1:
the series only contains a also
$\frac{1}{2}(2a+ (1-1)d)=a$ that proves the base case.
assume for some n=k
$a+(a+d)+ \cdots + (a+(k-1)d) = \frac{k}{2}(2a + (k-1)d)$ (1)
therefore adding a+kd to (1) yields :
$a+(a+d) \cdots + (a+(k-1)d) + (a+kd) = \frac{k}{2}(2a + (k-1) d) + (a+kd)$
$=(\frac{k}{2} (2a) + a) + (\frac{k(k-1)}{2}d +kd)$
$= (a(k+1) ) + (kd(\frac{k-1}{2} + 1))$
$= (\frac{k+1}{2}(2a) ) + (d(\frac{k(k+1)}{2} ))$
which is the statement for k+1.