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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 $T_n=a+(n-1)d$. 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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