Matryoshka Sequences

The sum of the $100$ first terms of $S_{100}$ is...

• the same as the sum of the first $200$ terms of $S_{99}$

• the same as the sum of the first $400$ terms of $S_{98}$

...

• the same as the sum of the first $100.2^{n}$ terms of $S_{100-n}$

...

• the same as the sum of the first $100.2^{99}$ terms of $S_1$ i.e. the sum of the first $100.2^{99}$ integers, which can be written as :

$\frac{100.2^{99}(1+100.2^{99})}{2} = 50(2^{99}(1+100.2^{99}) = 50(2^{99}+2^{99}.(1+99).2^{99}) = 50(2^{99}+2^{198}+99\times 2^{198})$

Firstly, we see that the sum of the first $k$ terms of $S_n$ is the same as the sum of the first $\frac{k}{2}$ terms of $S_{n+1}$ .

For example, the first $6$ terms of $S_1$ sum to $(1) + (2) + (3) + (4) + (5) + (6) = 21$ . Also, the first $3$ terms of $S_2$ sum to $(1+2) + (3+4) + (5+6) = 21$ .

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Therefore, $\text{Sum of the first } 100 \text{ terms of } S_{100} = \text{Sum of the first } (100 \cdot 2^1) \text{ terms of } S_{99}$ $\text{Sum of the first } 100 \text{ terms of } S_{100} = \text{Sum of the first } (100 \cdot 2^2) \text{ terms of } S_{98}$ $\text{Sum of the first } 100 \text{ terms of } S_{100} = \text{Sum of the first } (100 \cdot 2^3) \text{ terms of } S_{97}$ $\vdots$ $\text{Sum of the first } 100 \text{ terms of } S_{100} = \text{Sum of the first } (100 \cdot 2^{99}) \text{ terms of } S_{1}$

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Now we know that $\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$ So, $\displaystyle \sum_{k=1}^{100 \cdot 2^{99}} k = \frac{(100 \cdot 2^{99})(100 \cdot 2^{99}+1)}{2}$ $\displaystyle \sum_{k=1}^{100 \cdot 2^{99}} k = (50 \cdot 2^{99})(100 \cdot 2^{99}+1)$ $\displaystyle \sum_{k=1}^{100 \cdot 2^{99}} k = 50 \cdot (100 \cdot 2^{99} \cdot 2^{99}+ 2^{99})$ $\displaystyle \sum_{k=1}^{100 \cdot 2^{99}} k = 50 \cdot (100 \cdot 2^{198} + 2^{99})$ $\displaystyle \sum_{k=1}^{100 \cdot 2^{99}} k = 50(2^{99} + 2^{198} + 99 \times 2^{198})$