Christmas Streak 80/88: Limits on Geometry #1

Calculus Level 5

As shown above, there is a line segment with length 1. 1.

Define a mutation as doing the following:

For each segment left from the original line segment, divide it into three parts.

Make a square with the middle part, and erase that middle part.

T n {\rm T}_n is the shape obtained after n n mutation s, and l n l_{n} is the total length of the segments in T n . {\rm T}_n.

Find the value of ∑ n = 1 ∞ ( 4 − ∑ k = 1 n ( 3 − l k ) ) . \displaystyle \sum_{n=1}^{\infty}\left(4-\sum_{k=1}^{n}(3-l_k)\right).

This problem is a part of <Christmas Streak 2017> series .


The answer is 8.0.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Patrick Corn
Dec 27, 2017

I think you want ( 3 − l k ) (3-l_k) instead of ( 3 − l n ) (3-l_n) there, no?

Anyway, l k = ∑ i = 0 k ( 2 / 3 ) k = 1 − ( 2 / 3 ) k + 1 1 − 2 / 3 = 3 ( 1 − ( 2 / 3 ) k + 1 ) , l_k = \sum_{i=0}^k (2/3)^k = \frac{1-(2/3)^{k+1}}{1-2/3} = 3\left(1-(2/3)^{k+1} \right), so 3 − l k = 3 ( 2 / 3 ) k + 1 = 4 3 ( 2 / 3 ) k − 1 . 3-l_k = 3(2/3)^{k+1} = \frac43(2/3)^{k-1}. Then our sum is ∑ n = 1 ∞ ( 4 − 4 3 ∑ k = 1 n ( 2 / 3 ) k − 1 ) = ∑ n = 1 ∞ ( 4 − 4 3 1 − ( 2 / 3 ) n 1 − 2 / 3 ) = ∑ n = 1 ∞ ( 4 − 4 ( 1 − ( 2 / 3 ) n ) ) = ∑ n = 1 ∞ 4 ( 2 / 3 ) n = 4 2 / 3 1 − 2 / 3 = 8 . \begin{aligned} \sum_{n=1}^\infty \left( 4 - \frac43 \sum_{k=1}^n (2/3)^{k-1} \right) &= \sum_{n=1}^\infty \left( 4-\frac43 \frac{1-(2/3)^n}{1-2/3} \right) \\ &= \sum_{n=1}^\infty \left( 4-4(1-(2/3)^n) \right) \\ &= \sum_{n=1}^\infty 4(2/3)^n \\ &= 4 \frac{2/3}{1-2/3} = \fbox{8}. \end{aligned}

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