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Archive for January, 2010

AMC 8 Results for Texas

Posted in AMC 8, Competitions, tagged , on January 1, 2010 | Leave a Comment »


The AMC has published state level summaries for the 2009 AMC 8.  Texas did very well with several perfect scores.

 

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Telescoping Sums and Products (Solutions 1)

Posted in problems, tagged , , , on January 1, 2010 | Leave a Comment »


Here is the solution provided for this previous problem on Telescoping Sums and Products

Example 1.  Prove that

\displaystyle\sum^{n}_{k=1}k^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}

Example 1.  Solution

From the Binomial Theorem we have

\displaystyle(k+1)^5-k^5=5k^4+10k^3+10k^2+5k+1

thus

\displaystyle\sum^{n}_{k=1}[(k+1)^5-k^5]=(n+1)^5-1

It follows that

\displaystyle 5\sum^{n}_{k=1}k^4+10 \sum^{n}_{k=1}k^3+10 \sum^{n}_{k=1}k^2+5 \sum^{n}_{k=1}k+\sum^{n}_{k=1}1=(n+1)^5-1

Hence

\displaystyle5 \sum^{n}_{k=1}k^4=(n+1)^5-10\dfrac{n(n+1)(2n+1)}{6}-5\dfrac{n(n+1)}{6}-(n+1)

Yielding

\displaystyle\sum^{n}_{k=1}k^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}

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