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THUDandBLUNDER
Threefold Repetition

 Posted: Sun Sep 12, 2010 1:45 am    Post subject: 1 (1 + d) + 2(1 + 2d)r + 3(1 + 3d)r 2 + 4(1 + 4d)r 3 + ......... + n(1 + nd)r n-1 ....... when lrl < 1 (One I thought of myself.)Last edited by THUDandBLUNDER on Sun Sep 12, 2010 2:47 am; edited 2 times in total
Trojan Horse
Daedalian Member

 Posted: Sun Sep 12, 2010 2:26 am    Post subject: 2 Clarification needed: do you want the infinite sum, or just the sum to n terms? The way you've written it, it looks like you just want the sum to n terms. But given the title of the puzzle, and the fact that you said |r| must be less than 1, I'm guessing you want the infinite sum. (I'm too lazy to work it out either way. But still... )
THUDandBLUNDER
Threefold Repetition

Posted: Sun Sep 12, 2010 2:45 am    Post subject: 3

 Trojan Horse wrote: Clarification needed: do you want the infinite sum, or just the sum to n terms? The way you've written it, it looks like you just want the sum to n terms.

Yes, good point. Looks like I lost an ellipsis. Have amended it. Thanks.
.
ralphmerridew
Daedalian Member

 Posted: Sun Sep 12, 2010 10:14 am    Post subject: 4 S = sum(n = 1 to inf: n(1+nd)*r^(n-1)) rS = sum(n = 1 to inf: n(1+nd)*r^n) rS = sum(n = 2 to inf: (n-1)*(1+(n-1)d)*r^(n-1)) S - rS = 1*(1+d)*r^0 + sum(n = 2 to inf: (n(1+nd) - (n-1)*(1+(n-1)d)) r^(n-1)) S(1 - r) = 1+d + sum(n = 2 to inf: (n+nnd - (n-1)*(1+nd-d)) r^(n-1)) S(1 - r) = 1+d + sum(n = 2 to inf: (n+nnd - n -nnd+nd + 1+nd-d) r^(n-1)) S(1 - r) = 1+d + sum(n = 2 to inf: (1 + 2nd - d) r^(n-1)) S(1 - r)r = r(1+d) + sum(n = 2 to inf: (1 + 2nd - d) r^(n)) S(1 - r)r = r(1+d) + sum(n = 3 to inf: (1 + 2(n-1)d - d) r^(n-1)) S(1 - r) - S(1 - r)r = 1+d + sum(n = 2 to inf: (1 + 2nd - d) r^(n-1)) - (r(1+d) + sum(n = 3 to inf: (1 + 2(n-1)d - d) r^(n-1))) S(1 - r)^2 = 1+d + (1+4d - d)*r + sum(n = 3 to inf: (1 + 2nd - d) r^(n-1)) - r(1+d) - sum(n = 3 to inf: (1 + 2(n-1)d - d) r^(n-1)) S(1 - r)^2 = 1+d - r(1+d) + (1+3d)*r + sum(n = 3 to inf: ((1 + 2nd - d) - (1 + 2(n-1)d - d)) r^(n-1)) S(1 - r)^2 = 1+d-2rd + sum(n = 3 to inf: 2d r^(n-1)) S(1 - r)^2 = 1+d-2rd + 2d / (1-r) S = (1 + d - 2rd) / (1-r)^2 + 2d / (1-r)^3 Unless I made a mistake in the algebra, which is quite likely.
THUDandBLUNDER
Threefold Repetition

 Posted: Sun Sep 12, 2010 10:55 am    Post subject: 5 I think your first term is a bit out. Luckily, there is a quicker way: I obtained the given series by differentiating the following AGP term-by-term wrt r: 1 + (1 + d)r + (1 + 2d)r 2 + (1 + 3d)r 3 + .... So the sum to infinity of the series ought to be the differential coefficient of the AGP's sum to infinity. Hence we need to differentiate 1/(1 - r) + dr/(1 - r) 2 wrt r. This gives 1/(1 - r) 2 + d(1 + r)/(1 - r) 3 which equals [(d - 1)(r + 1) + 2]/(1 - r) 3 .
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