Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 16, 2011"

Latest revision as of 06:06, 16 June 2011

The equation for the convergent sum of an infinite geometric sequence is $S=\dfrac{a}{1-r}$ where a is the first term and r is the common ratio.

Therefore: $2=\dfrac{k}{1-n}$ and $3=\dfrac{n}{1-k}$

Transposing this and solving the simultaneous equations:

$2-2n=k$ $3-3k=n$

To find $n$ : $3-3(2-2n)=n$ $\dfrac{3}{5}=n$ To find $k$ : $2-2(\dfrac{3}{5})=k$ $\dfrac{4}{5}=k$

Therefore $k+n=\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{7}{5}=1\dfrac{2}{5}$

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 {{:AoPSWiki:Problem of the Day/June 16, 2011}}
 ==Solution==
-{{potd_solution}}
+The equation for the convergent sum of an infinite [[geometric sequence]] is <math>S=\dfrac{a}{1-r}</math> where ''a'' is the first term and ''r'' is the common ratio.
+Therefore:
+<cmath>2=\dfrac{k}{1-n}</cmath> and <cmath>3=\dfrac{n}{1-k}</cmath>
+Transposing this and solving the simultaneous equations:
+<cmath>2-2n=k</cmath>
+<cmath>3-3k=n</cmath>
+To find <math>n</math>: <cmath>3-3(2-2n)=n</cmath>
+<cmath>\dfrac{3}{5}=n</cmath>
+To find <math>k</math>: <cmath>2-2(\dfrac{3}{5})=k</cmath>
+<cmath>\dfrac{4}{5}=k</cmath>
+Therefore <math>k+n=\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{7}{5}=1\dfrac{2}{5}</math>