CMSC 250 Homework 7 Answers Fall 2001

1. 1. $a_{1} = 1^{1} ,\,\,a_{2}= 2^{2},\,\,a_{3}= 3^{3},\,\,a_{4}= 4^{4} ,\,\,a_{5}= 5^{5} ,\,\,a_{6}=6^{6}$

   2. $a_{1} = 1^{1},\,\,a_{2}= 2^{1/2},\,\,a_{3}= 3^{1/3},\,\,a_{4}= 4^{1/4} ,\,\,a_{5}= 5^{1/5},\,\,a_{6}= 6^{1/6}$

   3. $a_{1} = \frac{1}{2},\,\,a_{2}=\frac{2}{3} ,\,\,a_{3}=\frac{3}{4} ,\,\,a_{4}=\frac{4}{5} ,\,\,a_{5}=\frac{5}{6} ,\,\,a_{6}=\frac{6}{7}$

   4. $a_{1} =\frac{1}{3} ,\,\,a_{2}=\frac{1}{9},\,\,a_{3}=\frac{1}{27} ,\,\,a_{4}=\frac{1}{81} ,\,\,a_{5}=\frac{1}{243} ,\,\,a_{6}=\frac{1}{729}$

2. 1. $1^{1} + 2^{2}+ 3^{3}+ 4^{4}+ 5^{5}+ 6^{6} = 50069$

   2. $1^{1} +2^{1/2} +3^{1/3}+4^{1/4} +5^{1/5}+ 6^{1/6} = 7.99841$

   3. $\frac{1}{2} +\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7} =4.407142857 = \frac{617}{140}$

   4. $\frac{1}{3} +\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+ \frac{1}{729} = \frac{3 (1-\frac{1}{3}^{7})}{2}$

3. 1. $1^{1}* 2^{2}* 3^{3}* 4^{4} = 27648$

   2. $1^{1} *2^{1/2}* 3^{1/3}*4^{1/4} = 2.8845$

   3. $\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{4}{5} = \frac{1}{5}$

   4. $\frac{1}{3}\frac{1}{9}\frac{1}{27}\frac{1}{81}= \frac{1}{54049}=\frac{1}{3^{10}}$

4. 1. $\sum_{k =1}^{7}(-1)^{k+1}\frac{1}{k}$

   2. $\prod_{k=2}^{8}\frac{1}{k^{3}}$

   3. $\sum_{k = 1}^{8}\left(\frac{k}{k+1}\right)^{k+2}$

5. 1. Base Case:
      	$n=1$, $\sum_{i = 0}^{1} 2\left(\frac{1}{3}\right)^{i+1} = 2\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)^{2} =$
      	$\frac{6}{9}+\frac{2}{9} = \frac{8}{9} = 1 - \left(\frac{1}{3}\right)^{2}$
      Inductive Hypothesis:
      	$\sum_{i = 0}^{k-1} 2\left(\frac{1}{3}\right)^{i+1} = 1 -\left(\frac{1}{3}\right)^{k}$.
      Inductive Step:
      Show:
      	$\sum_{i = 0}^{k} 2\left(\frac{1}{3}\right)^{i+1} = 1 -\left(\frac{1}{3}\right)^{k+1}$.
      Proof:
      	$\sum_{i = 0}^{k} 2\left(\frac{1}{3}\right)^{i+1} = \sum_{i=0}^{k-1}2\left(\frac{1}{3}\right)^{i+1} + 2\left(\frac{1}{3}\right)^{k+1}$
      	$= 1 - \left(\frac{1}{3}\right)^{k} + 2\left(\frac{1}{3}\right)^{k+1},$ by the inductive hypothesis.
      	$= 1 - 3\left(\frac{1}{3}\right)^{k+1} + 2\left(\frac{1}{3}\right)^{k+1} = 1 - \left(\frac{1}{3}\right)^{k+1}.$ QED

   2. Base Case:
      	$n=1$, $4(1)-2 = 2(1)^{2}$.
      Inductive Hypothesis:
      	$2+6+10+...+(4(k-1)-2) = 2(k-1)^{2}.$
      Inductive Step:
      Show:
      	$2+6+10+...+(4k-2) = 2k^{2}.$
      Proof:
      	$2+6+10+...+(4k-2) = 2+6+10+...+(4(k-1)-2)+(4k-2).$
      	By the inductive hypothesis, $2+6+10+...+(4(k-1)-2) = 2(k-1)^{2}$,
      	so that $2+6+10+...+(4(k-1)-2)+(4k-2) = 2(k-1)^{2}+4k-2 =$
      	$(2k^{2}-4k+2)+(4k-2) = 2k^{2}.$ QED.

   3. Base Case:
      	$n=1$, $\prod_{i=1}^{1}\frac{i+1}{i} = \frac{2}{1} = 1 + 1 = n+1$.
      Inductive Hypothesis:
      	$\prod_{i=1}^{k-1}\frac{i+1}{i}=k$
      Inductive Step:
      Show:
      	$\prod_{i=1}^{k}\frac{i+1}{i}=k+1$
      Proof:
      	$\prod_{i=1}^{k}\frac{i+1}{i}=\frac{k+1}{k}\left(\prod_{i=1}^{k-1}\frac{i+1}{i}\right)$
      	By the induction hypothesis, $\prod_{i=1}^{k-1}\frac{i+1}{i}=k$,
      	which means that $\prod_{i=1}^{k}\frac{i+1}{i}=\frac{k+1}{k}(k) = k+1$. QED

   4. Base Case:
      	$n=6$, $n=6$, $6^{2}>3(6)+12$.
      Inductive Hypothesis:
      	$k^{2} > 3k+12$
      Inductive Step:
      Show:
      	$(k+1)^{2} > 3(k+1)+12$
      Proof:
      	$(k+1)^{2} = k^{2}+2k+1.$ $3(k+1)+12 = (3k+12)+3$.
      	By the inductive hypothesis, $k^{2} > 3k+12$.
      	Because $k\geq 6$, $2k+1 > 3$, so $k^{2}+(2k+1) > (3k+12)+3$,
      	which means that $(k+1)^{2} > 3(k+1)+12$. QED

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