Sum of iDate: 03/23/2002 at 09:57:56 From: Julie Gal Subject: Summation If the sum as i goes from 1 to n of 2^i is 2^n -1, what is the sum as i goes from 1 to n of 3^i ? Thanks. Julie Gal Date: 03/23/2002 at 12:22:52 From: Doctor Jubal Subject: Re: Summation Hi Julie, Thanks for writing Dr. Math. I started by writing down the first few sums, and here's what I got n 3^n sum (3^i) 0 1 1 1 3 4 2 9 13 3 27 40 4 81 121 5 243 364 6 729 1093 It appears to me, that in each case, the sum is one less than half of 3 to the next highest power: 1 = (3-1)/2 4 = (9-1)/2 13 = (27-1)/2 40 = (81-1)/2 etc. So, let's see if we can prove this: sum from 1 to n of 3^i = (3^(n+1)-1)/2. We'll do it by mathematical induction. First, the basis step. For n=0, 1 = (3^1 - 1)/2 = 1. Now, the induction step. If we suppose that the statement is true for some value of n, does it necessarily follow that it is also true for the next highest value of n? Since the nth sum is (3^(n+1) - 1)/2, if we add the next term to it, we get (3^(n+1) -1)/2 + 3^(n+1) = (3/2)*3^(n+1) - 1/2 = (3* 3^(n+1) - 1) / 2 = (3^(n+2) - 1)/2 Which is exactly what our proposed formula would predict for the value of the sum from i=1 to n+1. So, if the formula is true for some value of n, it must also be true for all greater values of n. However, we've already observed that it is true for n=0, so we can conclude that it is also true for all postive n. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/ Date: 03/23/2002 at 22:10:44 From: Julie Gal Subject: Summation Dear Dr Jubal: Thank you for your solution to my question. I followed your proof. Your summation is from i = 0 to n instead of i = 1 to n, which of course also includes the number 1. My math 12 honors class is studying sequences and series and that would be a great extra credit question to have them work on. Sincerely, Julie Gal |
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