Summing A Very Interesting Infinite Series

[@Qermaq]

This was a fun one. Hey, here's a fairly easy one for the holidays: on the first day of Christmas your true love gives to you 1 present a. (n = 1 ==> a) On the second day of Christmas your true love gives to you 2 presents b and 1 present a. (n = 2 ==> 2b + a) On the third day of Christmas your true love gives to you 3 presents c, 2 presents b and 1 present a. (n = 3 ==> 3c + 2b + a) (1) In total, how many presents will your true love give to you on day n? (2) How many presents will your true love have given you in total from days 1 to n?

[@schwidola3549]

The second method is a little bit of black magic, because you would have to foresee that deriving would produce an interesting result. Good insight, but the first method is more satisfying, because you realize that you have a nested geometric series, and the very thought of decomposing a sum into other sums is very useful for other problems

[@scottleung9587]

Good point!

[@robertcotton8481]

I agree with wolframalpha jyst glance thought we were doing fibanochie

[@bobkurland186]

I did method 1

[@dan-florinchereches4892]

I think the sum is i*x^(3i-1) i>=1 So multiplying by 3 and integrating results in the series X^3i geometric series which has Sum(1..n) = (1-x^(3n+1))/(1-x^3) And infinite sum is 1/(1-x^3) when |x|

[@scottleung9587]

Nice!

[@SyberMath]

Glad you liked it!

[@parimalpandya9645]

क्या रामानुजन और महाबलेनियस का हाउस नंबर प्रोब्लम का एक और सुझाव है

[@dwm1943]

I did method 1, or do I mean 2? Without the differentiation. Easy, but fun. |x| < 1.

[@davidmitchell3881]

put x = -1 and the sum is rather interesting 😂

[@MrGeorge1896]

x² + 2x⁵ + 3x⁸ + ... = x² ( 1 + 2x³ + 3x⁶ + ...) substitute x³ with y: x² ( 1 + 2y + 3y² + ...) but we know the second factor is equal to 1 / (1 - y)² x² / (1 - y)² = x² / (1 - x³)² = (x / (1 - x³))²

[@MohammadElmi]

How do you know the second factor? This is how I calculated it: Suppose 1 + 2y + 3y^2 + ... = s. Integrating both sides gives: y + y^2 + y^3 + ... = S (supposing s = dS/dy). Using the formula for sum of infinite series: S = y/(1-y). Differentiate and you get: s = 1/(1-y)^2. Do you have a simpler method?

[@MrGeorge1896]

@@MohammadElmi This series is so common that I consider it as well known. 😀 But we can also derive it as follows: (1 + 2y + 3y² +....) = 1 * (1 + y + y² +....) + y * (1 + y + y² +....) + y² * (1 + y + y² +....) +... = (1 + y + y² +....) * (1 + y + y² +....) = 1 / (1 - y) * 1 / (1 - y) = 1 / (1 - y)²

[@ahmadalizade-h2k]

The second is not true. These equations only hold |x|

[@agytjax]

Haven't you heard what Shakespeare has said : "It doesn't matter which method came first, it would suck just the same !"😇

[@FisicTrapella]

Shakespeare said that when he realized the second method came first 😄

[@forcelifeforce]

Stop spamming your post, you rude *a-hole!*

[@seanfraser3125]

(x/(1-x^3))^2

[@guyhoghton399]

This is my second comment, but I've decided to post it first. 🙂

[@SyberMath]

Haha! Noooo, you can’t do that! 🤪😁

[@guyhoghton399]

@@SyberMath 😀

[@agytjax]

Why does this guy always annoys us with his BS : "I am gonna start with the 2nd method first" 😀😀 ? Haven't you heard what Shakespeare has said : "It doesn't matter which method came first, it would suck just the same !"

[@guyhoghton399]

_Let S = x² + 2x⁵ + 3x⁸ + 4x¹¹ + ..._ _x³S = x⁵ + 2x⁸ + 3x¹¹ + ...._ ∴ _S - x³S = x² + x⁵ + x⁸ + x¹¹ + ..._ ⇒ _(1 - x³)S = x² / (1 - x³) if |x| < 1_ ⇒ *_S = x² / (1 - x³)²_*

[@SyberMath]

Pretty good!