Ex (1): The number of bacteria in a colony doubles every hour. If a colony begin with five bacteria, how many will be present in n hours?
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{a_n} = 2\;{a_{n - 1}}\;Where\;\;{a_n} = \# \;\;of\;bactiria\;at\;time\;n,\;with\;intial\;condintion\;{a_0} = 5\\
\Rightarrow {a_n} = 2\;{a_{n - 1}}\;,\;\;{a_0} = 5\\\\
Observe\;that\;\;\;\;{a_1} = 2\;{a_0} = 2 * 5 = 10\\
{a_2} = 2\;{a_1} = 2 * 10 = 20\;\\
{a_3} = 2\;{a_2} = 2 * 20 = 40
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Ex\left( 2 \right):\;{a_n} = 2\;{a_{n - 1}} - \;{a_{n - 2}}\;,\;\;\;\;{a_0} = 0\;,{a_1} = 3\\\\
1 - Find\;{a_2},\;{a_3},\;\;{a_4}\;.\\\\
\Rightarrow \;\;{a_2} = 2\;{a_1} - {a_0} = 2\left( 3 \right) - 0 = 6\\
\;{a_3} = 2\;{a_2} - {a_1} = 2\left( 6 \right) - 3 = 9\\
\;{a_4} = 2\;{a_3} - {a_2} = 2\left( 9 \right) - 6 = 12\\\\
2 - Determine\;whetther\;\;{a_n} = 3n,\;is\;a\;solution\;for\;this\;recurrence\;relation\;.\\\\
\Rightarrow plug\;in\;{a_n} = 3n\;,\;in\;the\;recurrence\;relation\;,\\
\;{a_n} = 2\;{a_{n - 1}} - \;{a_{n - 2}}\\
3n\;does\;it\;equal\;2\left( {3n - 3} \right) - \left( {3n - 6} \right)\\
\left( {6n - 6} \right) - \left( {3n - 6} \right) = 3n\\
Thus\;\;{a_n} = 3n\;\;is\;a\;solution\;of\;the\;recurrence\;relation.\\\\
3 - Is\;{a_n} = {2^n},\;\;a\;solution\;for\;this\;recurrence\;relation\;?\;\\\\
\Rightarrow \;\;{a_n} = 2\;{a_{n - 1}} - \;{a_{n - 2}}\\
{a_n} = {2^n} \Rightarrow {2^n}does\;it\;equal\;2 * {2^{n - 1}} - {2^{n - 2}}\\
{2^n} - {2^{n - 2}}\\
{2^n}\; \ne {2^n} - {2^{n - 2}}\\
Thus\;\;{a_n} = {2^n}\;\;is\;a\;solution\;of\;the\;recurrence\;relation.
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