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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 $\displaystyle =\ a(r^n-1)$ $\displaystyle \therefore\ \sum_{k=1}^n\,T_k\ =\ \frac{a(r^n-1)}{r-1}$ Tags arithmetic, geometric, progression Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jiasyuen Algebra 5 April 23rd, 2014 01:42 PM rnck Real Analysis 3 November 18th, 2013 09:19 PM jareck Algebra 3 July 6th, 2012 07:38 AM Algebra 2 April 5th, 2012 10:46 PM Francis410 Algebra 1 March 22nd, 2011 08:02 AM

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