I'm trying to prove the following equation by induction, but my base case isn't working. 
for all n>2.
For my base case I did n=3, and on the LHS I got 8/9 and the RHS I got 2/3.
Helppp.
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ematth7
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You miscalculated. $(1-\frac{1}{2^2})(1-\frac{1}{3^2}) = \frac{3}{4} \cdot \frac{8}{9} = \frac{24}{36} = \frac{2}{3}$ – Falko Sep 03 '15 at 13:00
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Base case gives 2/3 LHS and RHS – frabala Sep 03 '15 at 13:00
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For $n=3$ your lhs is $(1-\frac 14)(1-\frac 19)=\frac 34,\frac 89=\frac 23$. – lulu Sep 03 '15 at 13:01
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OHHHH. I get it now. I was just plugging 3 in for (1-1/n^2) not also using what comes before it. – ematth7 Sep 03 '15 at 13:02
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And then quick question. To prove using induction, would I multiply (n+1) to both sides of the equation? – ematth7 Sep 03 '15 at 13:20
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The left hand side is
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)=\frac{3}{4}\frac{8}{9} = \frac23$$
and so is the right hand side:
$$\frac{3 + 1}{2\cdot 3} = \frac46=\frac23$$
5xum
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And then quick question. To prove using induction, would I multiply (n+1) to both sides of the equation or add (n+1)? – ematth7 Sep 03 '15 at 13:34
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@ematth7 You would change all values of $n$ to $n+1$. So, on the left, you would get one more factor in the multiplication, and on the right, you would get $$\frac{(n+1) + 1}{2\cdot (n+1)}$$ – 5xum Sep 03 '15 at 14:40