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May 1st, 2014, 01:03 AM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  Arithmetic Progression and Geometric Progression
How to proof those formula in these two progressions?

May 1st, 2014, 04:40 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
(1) Arithmetic progression with first term $a$ and common difference $d$. The $n$th term is $T_n=a+(n1)d$. Thus $$\sum_{k=1}^n\,T_k$$ $\displaystyle =\ \sum_{k=1}^n\,[a+(k1)d]$ $\displaystyle =\ \sum_{k=1}^n\,a \, + \, d\sum_{k=1}^n\,(k1)$ $\displaystyle =\ na+d\frac{(n1)n}2$ $\displaystyle =\ \frac n2\left[2a+(n1)d\right]$ (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 $T_n=ar^{n1}$. Thus $(r1)T_n=a(r^nr^{n1})$ and $$\sum_{k=1}^n\,(r1)T_k$$ $\displaystyle =\ a\sum_{k=1}^n\,(r^kr^{k1})$ $\displaystyle =\ a(r^nr^0)$ by the method of telescoping $\displaystyle =\ a(r^n1)$ $\displaystyle \therefore\ \sum_{k=1}^n\,T_k\ =\ \frac{a(r^n1)}{r1}$ 

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arithmetic, geometric, progression 
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