Sequences and series *: Rekenkundige rijen *
Arithmetic series
An arithmetic sequence has the form #a_k = ck+d# with #k\in\mathbb{N}# and #c#, #d\in\mathbb{R}#.
Hence the terms #a_k# have the values of a polynomial of degree one (the function that adds the value #cx+d# to #x#) in the positive integers.
Characteristic for an arithmetic sequence is the fact that the difference between two subsequent numbers is constant, being #c#.
An arithmetic series originates by taking the summand of an arithmetic sequence:\[b_n=\sum_{k=1}^na_k = \sum_{k=1}^n\left(ck+d\right)=c\left(\sum_{k=1}^nk\right)+n\cdot d\tiny.\]
We will show with full induction that the formula #\displaystyle\sum_{k=1}^{n} k = \dfrac{1}{2}n(n+1)# is true for all #n\ge1#.
If #n=1#, then the left hand side is #1# and the right hand side is #\dfrac{1}{2}\cdot1\cdot 2#, which is equeal to #1#. So the formula is correct for #n=1#.
Consider that the formula is correct for a certain value #n\ge1#. We will show that the formula is also correct for #n+1#. On the left hand side
\begin{eqnarray*}\sum_{k=1}^{n+1} k &=&\left( \sum_{k=1}^{n} k\right) + n+1=\dfrac{1}{2}n(n+1) + (n+1) \\ &=& \dfrac{1}{2}\left[n(n+1)+2(n+1)\right]= \dfrac{1}{2}(n+1)(n+2)\tiny.\end{eqnarray*}
At the end we see the exact formula for #n+1# in stead of #n#. The rule follows from the principle of full induction.
Because, the formula gives\[\sum_{k=1}^{11}a_k = 5\left( \sum_{k=1}^{11} k\right) + 11\cdot (-1) = \dfrac{5}{2}\cdot11(11+1) -11 = 319\tiny.\]
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