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quiz 3

# quiz 3 - l‘ lJI‘OVG by 1nduct10‘n that IOI‘ all n 2...

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Unformatted text preview: l‘) lJI‘OVG by 1nduct10‘n that IOI‘ all n 2 4, 671 < m. " 2) Prove by induction that for all n 2 1 9+(9x10)+(9><100)+~-+(9><10"‘1)=1O"—1. Gagem2 ’PM) wwhu ansqwslw ‘2 M], JEN?“ \NWM NWYWs ; ASSN vb“ «hm Jedi 8“ “EH ‘ a . \ c ‘ Mu w SKP , W Vm) awn) «mm WW) \SWWJWW) 3 Wm; gumm L (rm)! 3n *3 <m~»\1v\)m (2mm . ~ ) ‘ “:3 WM» 11h) 3m (Y): QM 3®€8M 3W3 ( ﬂ’, +3. "301“) L (WXH—dn-ﬂhh +3 MKm/O “‘“3 mm “a” . ) so M m I ‘ - 5m“) < (WHHXWIIM) . 8 1’7 MAYER“? 1211 \$4211” “4:35 . ammme a .7 Cr I .. SlYlCX ﬁLMNSqu/Bmwatypéﬂn) ’pmbpmu) R" m w pnmm mmmmrmex \Mvc’vm pm J M WW h) (94m; Sthwu “WM (4+ Lqmovmmﬁ. L(q»n:‘")/:;a€ mxxgxwu)“ 0r ‘IO'l . ,._ . - — =01 J WHEN 02/ Wm“ WWW) ; Assam pm) \2 W W 3“ \AH P W , . \ m 3e? mm wmpmlu) Wmm \s WW) 8 NW q+(‘?MO)L.N+WW® ‘t IOVHI *(qx\0““3*(olx\0“') I ’ ’/ mam mm m) Ot+(qx\o)*..+(°m0“") ‘1 +<q>403+ . *(q *\0“")* (‘1 n0“ : 110“ DW 740“ ﬁg“); 3.. W ~ :WH' ‘+10'\(01x\6“ “LG I) I “OM . / ‘ ~ (“Ml-«47M. 0.! Jim WK em‘o Wyn.) Nlnl\\ nlmﬁ :. '«Iu 0. 3.41514 ~ /‘ J a“) 1k xi“ v1“ Math 300, Fall 2011 \A Quiz 3 Solutions 1) Prove by induction that for all n 2 4, 3n < n!,. /Let P(n) be the statement ‘For all n 2 4, 3n < n!’ JBase case: P(4) states that 12 < 24 which is true, but note that P(3) is false. \/ Inductive hypothesis: Assume that for some n 2 4, 3n < n!. Inductive step: We must show that P(n) => P(n + 1), where P(n + 1) is the statement 3n+3 < (n+1)!. . By the inductive hypothesis, we have 3n < n!. Thus 3n+3 <nl+3 < nl+nl <.n.n!+n! =nl(n+1) = (n+ 1)! Note that the second inequality is true since n 2 4. So P(n) : P(n+ 1) for all n 2 4, so by the Principle of Mathematical Induction, P(n) is true for all n 2 4. \_ 2) Prove by induction that for all n 2 1, 9+ (9 X 10) + (9 x 100) + - - - + (9 X lOn‘1)= 10" — 1. {Let P(n) be the statement ‘for all n 2 1, 9+ (9 x 10) + (9 X 100) + - v - + (9 X 10"‘1)= 10" — 1.’ ‘/Base case: LHS of P(l) = 9 X 100 = 9, RHS of P(l) = 101 — 1 = 9=LHS, so P(l) is true. /1 nductive hypothesis: Assume that for some n 2 1, 9+(9x10)+(9x100)+---+(9X10”"1)=10”—1. Inductive step: We must show that P(n) => P(n + 1), where P(n + 1) is the statement 9+(9X10)+(9x100)+...+(9x10n)=10n+1_1. / J Adding (9 x 10”) to both sides of P(n) we obtain 9+(9x10)+(9x100)+---+(9><10”‘1)+(9><10”)=10”—1+(9><10"). The RHS is ~ ,1. Dale Smiliﬁcamvx '10" — 1 + 9 x 10”) = 10”(1 +9) — 1 = 10”(10)— 1 = 10”+1 — 1 D“ L IO“ 1 }Un ma” which is the LHS of P(n + 1). So P(n) ==> P(n + 1) for all n 2 1, so by the Principle of Mathematical Induction, P(n) is true for all n 2 1. 1 ...
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quiz 3 - l‘ lJI‘OVG by 1nduct10‘n that IOI‘ all n 2...

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