Past HSC mathematical induction q's (2 Viewers)

puffynutty · Jan 19, 2009

1. Use mathematical induction to prove that for all integers n≥3
(1-2/3)(1-2/4)(1-2/5)...(1-2/n) = 2/n(n-1)

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ontVertAlignInTxbx/> <w:Word11KerningPairs/> <w:CachedColBalance/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> <m:mathPr> <m:mathFont m:val="Cambria Math"/> <m:brkBin m:val="before"/> <m:brkBinSub m:val="--"/> <m:smallFrac m:val="off"/> <m:dispDef/> <m:lMargin m:val="0"/> <m:rMargin m:val="0"/> <m:defJc m:val="centerGroup"/> <m:wrapIndent m:val="1440"/> <m:intLim m:val="subSup"/> <m:naryLim m:val="undOvr"/> </m:mathPr></w:WordDocument> </xml><![endif]--><style>  </style> 2. i) show that for all positive integers n
x[(1+x)<sup>n-1</sup>+(1+x)<sup>n-2</sup>+...+(1+x)<sup>2</sup>+(1+x)+1]=(1+x)<sup>n</sup>-1
<o></o>
ii) Hence show that for 1≤K≤n
n-1 n-2 n-3 K-1 n
+ + + ...+ =
K-1 K-1 K-1 K-1 K
<o></o><o></o>
iii) Show that n (n-1) = (k+1) ( n )
k k+1
<o></o>
IV) By differentiating both sides of the identity in (i), show that for 1≤K≤n,
<o></o>
(n-1) (n-2) + (n-2)(n-3) +...+K(K-1) =K ( n )
K-1 K-1 K-1 K+1

<o>(THE NUMBERS UNDERNEATH GO WITH THE BRACKETS ABOVE THEM THEY ARE NOT DIVISIONS THOUGH)
</o>

independantz · Jan 19, 2009

puffynutty said:
1. Use mathematical induction to prove that for all integers n≥3
(1-2/3)(1-2/4)(1-2/5)...(1-2/n) = 2/n(n-1)<o>
</o>

For n=3,

LHS= (1-2/3)=1/3
RHS=2/(3(2))=1/3=LHS

Assume true for n=k
i.e. assume: (1-2/3)(1-2/4)(1-2/5)...(1-2/k)=2/(k(k-1))

Prove True for n=k+1,
i.e. prove (1-2/3)(1-2/4)(1-2/5)...(1-2/k)(1-2/(k+1))=2/((k+1)k)

LHS=(1-2/3)(1-2/4)(1-2/5)...(1-2/k)(1-2/(K+1))
=2/(k(k-1))x(1-2/(k+1)), by assumption
=2/(k(k-1))x(k-1/k+1)
=2k-2/(k(k+1)(k-1))
=2/(k(k+1))
=RHS

*insert conclusion*

puffynutty said:
2. i) show that for all positive integers n
x[(1+x)n-1+(1+x)n-2+...+(1+x)2+(1+x)+1]=(1+x)n-1

If n=1,
LHS=x(1+x)⁰=x
RHS=(1+x)¹-1=x=LHS
therefore true for n=1

Assume true for n=k,where k is a positive integer
i.e. assume x[(1+x)k-1+(1+x)k-2+...+(1+x)2+(1+x)+1]=(1+x)^k-1

Prove true for n=k+1,
i.e. prove x[ (1+x)^k+(1+x)k-1+(1+x)k-2+...+(1+x)2+(1+x)+1]=(1+x)^k+1-1

LHS=x[ (1+x)^k+(1+x)k-1+(1+x)k-2+...+(1+x)2+(1+x)+1]
=x (1+x)^k+x[(1+x)k-1+(1+x)k-2+...+(1+x)2+(1+x)+1]
=x (1+x)^k+(1+x)^k-1
=(1+x)^k(x+1)-1
=(1+x)^{k+1^{-1

=RHS

*conclusion*}}

puffynutty · Jan 20, 2009

thanks it really helped

Past HSC mathematical induction q's (2 Viewers)

puffynutty

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independantz

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puffynutty

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