Induction Questions (1 Viewer)

rand_althor · Aug 5, 2015

Was looking through the 3U marathon and saw this question. Can someone help?

FrankXie said:
$$ Prove by mathematical induction that $ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\codts+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+ \frac {1}{2n}.$

Drongoski · Aug 5, 2015

LHS = $(1-\frac{1}{2}) + (\frac {1}{3} - \frac {1}{4}) + (\frac {1}{5} - \frac {1}{6}) = \dots + (\frac {1}{2n-1} - \frac {1}{2n})$

For n=1, LHS = 1/2 = RHS (so true for n=1)

Assume true for n = k >= 1

i.e. $(1 - \frac {1}{2}) +(\frac {1}{3} - \frac {1}{4}) + \cdots + (\frac {1}{2k-1} - \frac {1}{2k}) = \frac {1}{k+1} + \frac {1}{k+2} + \cdots + \frac {1}{2k} \\ \\ \therefore (\frac {1}{1} - \frac {1}{2}) + (\frac {1}{3} - \frac {1}{4}) + \cdots + (\frac {1}{2k-1} - \frac {1}{2k}) + (\frac {1}{2k+1} - \frac {1}{2k+2}) \\ \\ = \frac {1}{k+1} + \frac {1}{k+2} + \cdots \frac {1}{2k} + \frac {1}{2k+1} - \frac {1}{2k+2} \\ \\ = \frac {1}{k+2} + \frac {1}{k+3} + \cdots + \frac {1}{2k} + \frac {1}{2k+1} - \frac {1}{2k+2}+ \frac {1}{k+1}\\ \\ = \frac {1}{k+2} + \frac {1}{k+3} + \cdots + \frac {1}{2k} + \frac {1}{2k+1} + \frac {1}{2k+2}$

Therefore if true for n = k, also true for n = k+1

etc etc etc

Note

$\frac {1}{k+1} - \frac {1}{2k+2} = \frac {1}{2k+2}$

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Induction Questions (1 Viewer)

rand_althor

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Drongoski

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