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May 1st, 2014, 01:03 AM   #1
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Arithmetic Progression and Geometric Progression

How to proof those formula in these two progressions?
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May 1st, 2014, 04:40 AM   #2
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Math Focus: Abstract algebra
(1) Arithmetic progression with first term $a$ and common difference $d$.

The $n$th term is $T_n=a+(n-1)d$. Thus
$$\sum_{k=1}^n\,T_k$$
$\displaystyle =\ \sum_{k=1}^n\,[a+(k-1)d]$

$\displaystyle =\ \sum_{k=1}^n\,a \, + \, d\sum_{k=1}^n\,(k-1)$

$\displaystyle =\ na+d\frac{(n-1)n}2$

$\displaystyle =\ \frac n2\left[2a+(n-1)d\right]$


(2) Geometric progression with first term $a$ and common difference $r$.

Case 1: $r=1$
In this case the progression is a constant sequence consisting of $a$ in every term so the sum to $n$th term is $na$.

Case 2: $r\ne1$
The $n$th term is $T_n=ar^{n-1}$. Thus $(r-1)T_n=a(r^n-r^{n-1})$ and
$$\sum_{k=1}^n\,(r-1)T_k$$
$\displaystyle =\ a\sum_{k=1}^n\,(r^k-r^{k-1})$

$\displaystyle =\ a(r^n-r^0)$ by the method of telescoping

$\displaystyle =\ a(r^n-1)$

$\displaystyle \therefore\ \sum_{k=1}^n\,T_k\ =\ \frac{a(r^n-1)}{r-1}$
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