Algebra #18

by sjaelee, Aug 28, 2011, 6:36 PM

Quote:
Determine the value of

\[ S =\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\cdots+\sqrt{1+\frac{1}{1999^{2}}+\frac{1}{2000^{2}}}. \]

We start by plugging in $n=1$:

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+\frac{1}{n^2}+\frac{1}{(n+1)^2}$. Expanding messily,

$\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2}}=\sqrt{\frac{(n^2+n+1)^2}{(n(n+1))^2}}=$

$\frac{n^2+n+1}{n^2+n}$. I looked at it this way:

$\frac{a}{n}+\frac{b}{n+1}=\frac{n^2+n+1}{n^2+n}$. This means

$(a+b)n+a=n^2+n+1$. Letting $a=1$ and $b=n$ gave a solution (yay!), so

$\frac{1}{n}+\frac{n}{n+1}+\frac{1}{n+1}+\frac{n+1}{n+2}+...+\frac{1}{n+1998}+\frac{n+1998}{n+1999}$.

Note that evey term discluding the first and last terms going up can be paired to equal one, so

There are (every number counted twice expect $1,2000$):$2000*2-2=3998$ terms, so $3998-2=3996$ that can be paired for one, so $3996/2=1998$. Then,

$1998+\frac{1}{n}+\frac{n+1998}{n+1999}$ is desired. $n=1$, s

$1998+1+\frac{1999}{2000}=2000-\frac{1}{2000}$

$=\frac{3999999}{2000}$.
This post has been edited 1 time. Last edited by djmathman, Aug 28, 2011, 6:38 PM

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very interesting, i didn't know you could simplify an unparsable latex formula as such.

by tc1729, Aug 28, 2011, 6:37 PM

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I have pmed dj

by sjaelee, Aug 28, 2011, 6:38 PM

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Fixed. I pm'ed sjaelee.

Btw, is there a nickname you would like to be called? Or do you prefer sjaelee?

by djmathman, Aug 28, 2011, 6:42 PM

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hmmm...
sjaelee is my school e-mail adress
maybe?

by sjaelee, Aug 28, 2011, 6:50 PM

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