Pre-Calculus Pre-Calculus Math Forum

 December 26th, 2017, 03:14 PM #11 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 I am late to this thread but $\displaystyle \sum_{j=k}^n \text { expression in j and n }$ means to loop through, incrementing j by 1 each time, starting with j = k and ending with j = n, to calculate the value of the expression in each iteration of the loop, and to add that value to a cumulative sum. $\displaystyle \sum_{j=2}^5 nj^2$ is computed as follows Initialize sum = 0 First iteration j = 2 so $nj^2 = 5 * 2^2 = 20 \text { and } 0 + 20 = 20.$ Second iteration j = 3 so $nj^2 = 5 * 3^2 = 45 \text { and } 45 + 20 = 65.$ Third iteration j = 4 so $nj^2 = 5 * 4^2 = 80 \text { and } 80 + 65 = 145.$ Last iteration j = 5 so $nj^2 = 5 * 5^2 = 125 \text { and } 125 + 145 = 270.$ Summation notation is a highly abbreviated way to explain a procedure. Nothing fancy about it. December 28th, 2017, 03:32 PM   #12
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 Originally Posted by R_W_B . . . . . < cut > . . . . . And lastly when I attempted to plug $\sum \limits_{n=0}^\infty~\dfrac{(2n)!}{2^{2n}(n!)^2(2n +1)}x^{2n+1}$ into an online equation calculator, it gave this: $[ \sum \limits_{n=0}^\infty~\dfrac{(2n)!}{2^{2n}(n!)^2(2n +1)} ]x^{2n+1}$ but (to me) that would give an errant answer for any iterations of n = greater than 1, since it seems to separate the x^(2n+1) from the Sum. But maybe that calculator knows more than me (?) at this point.
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 Originally Posted by JeffM1 I am late to this thread but . . . . . .
Thanks for the reply, just now getting back on. Yes I understand all that now; my only confusion at this point is (shown above) when I plug into some equation calculators and hit enter, they put the parenthesis around the Sigma but not including the x^(2n+1) and this seems to me to exlude it from all iterations. Just a curiosity of why they do that or would that be wrong?

Last edited by skipjack; December 28th, 2017 at 04:02 PM. December 28th, 2017, 04:07 PM #13 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 Some calculators may be more flexible than others as to how a summation is entered. Tags function, inverse, question 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 WWRtelescoping Calculus 2 September 2nd, 2014 04:57 AM deSitter Algebra 4 April 10th, 2013 01:17 PM razzatazz Trigonometry 12 March 5th, 2013 11:08 PM annakar Calculus 3 January 11th, 2013 06:07 AM jaredbeach Algebra 1 November 17th, 2011 11:58 AM

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