2013 Mathcounts Basic Concept Review: Some Common Sums/numbers

These are some common sums that appear on Mathcounts often.

\(1+2+3+\ldots +n=\dfrac {n\left( n+1\right) } {2}\)

 \(2+4+6+\ldots .2n=n\left( n+1\right)\)

\(1+3+5+\ldots .\left( 2n-1\right) =n^{2}\)

The above are all arithmetic sequences.

The sum of any arithmetic sequence is average times terms (how many numbers).
Besides, the mean and median are the same in any arithmetic sequence.
Combining these knowledge, along with distributive rules some times
(case in point, sum of multiples of n, etc...) will expedite the calculation.

The "nth" triangular number is \(\dfrac {n\left( n+1\right) } {2}\)

The sum of the first n triangular numbers is \(\dfrac {n\left( n+1\right) \left( n+2\right) } {6}\).

 \(1^{2}+2^{2}+3^{2}+\ldots +n^{2}\) = \(\dfrac {n\left( n+1\right) \left( 2n+1\right) } {6}\)

\(1^{3}+2^{3}+3^{3}+\ldots +n^{3}=\)\(\left[ \dfrac {n\left( n+1\right) } {2}\right] ^{2}\)

Harder Geometry Question from National Mathcounts

Question: Similar to a 96 National Mathcounts question: Circles of the same colors are congruent and tangent to each other. What is the ratio of the area of the largest circle to the area of the smallest circle?

Solution :
The most difficult part might be to find the leg that is "2R - r".
R and r being the radius of the median and the smallest circle.

Using Pythagorean theorem, you have \(\left( R+r\right) ^{2}= R^{2}+\left( 2R-r\right) ^{2}\)
\(R^{2}+2Rr+r^{2}=R^{2}+4R^{2}-4Rr+r^{2}\)
Consolidate and you have 6Rr = \(4R^{2}\)\(\rightarrow\) 3r = 2R \(\rightarrow\) r =\(\dfrac {2R} {3}\) The ratio of the area of the largest circle to the area of the smallest circle is thus \(\rightarrow\)\(\dfrac {\left( 2R\right) ^{2}} {\left( \dfrac {2R} {3}\right) ^{2}}\) = 9 .