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 \(\dfrac {n\left( n+1\right) \left( n+2\right) } {6}\).
\(1^{2}+2^{2}+3^{2}+\ldots +n^{2}\) = \(\dfrac {n\left( n+1\right) \left( 2n+1\right) } {6}\)
\(1^{3}+2^{3}+3^{3}+\ldots +n^{3}=\)\(\left[ \dfrac {n\left( n+1\right) } {2}\right] ^{2}\)
Thursday, January 10, 2013
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