# arithmetic progression

The sum of terms of an arithmetic progression can be computed using Gauss’s trick:

$\displaystyle S$ $\displaystyle=\makebox[70.0pt]{(a_{1}+0)}+\makebox[70.0pt]{(a_{1}+d)}+% \cdots+\makebox[70.0pt]{(a_{1}+(n-2)d)}+\makebox[70.0pt]{(a_{1}+(n-1)d)}$ $\displaystyle+\underline{S\vphantom{\makebox[70.0pt]{(a_{1}+(n-1)d)}}}$ $\displaystyle\underline{{}=\makebox[70.0pt]{(a_{1}+(n-1)d)}+\makebox[70.0pt]% {(a_{1}+(n-2)d)}+\cdots+\makebox[70.0pt]{(a_{1}+d)}+\makebox[70.0pt]{(a_{% 1}+0)}}$ $\displaystyle 2S$ $\displaystyle=\makebox[70.0pt]{(2a_{1}+(n-1)d)}+\makebox[70.0pt]{(2a_{1}+(n% -1)d)}+\cdots+\makebox[70.0pt]{(2a_{1}+(n-1)d)}+\makebox[70.0pt]{(2a_{1}+(% n-1)d)}.$

We just add the sum with itself written backwards, and the sum of each of the columns equals to $(2a_{1}+(n-1)d)$. The sum is then

 $S=\frac{(2a_{1}+(n-1)d)n}{2}.$
Title arithmetic progression ArithmeticProgression 2013-03-22 13:39:00 2013-03-22 13:39:00 bbukh (348) bbukh (348) 10 bbukh (348) Definition msc 00A05 msc 11B25 MulidimensionalArithmeticProgression SumOfKthPowersOfTheFirstNPositiveIntegers