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# Do you understand?

Is $$2 + 2^2 + 2^3 + 2^4 + ... + 2^{10}$$ divisible by $$3$$?

It seems undetermined, because there were no threes, but let's try to pair it, extract it to find the threes.

$$A = (2 + 2^2) + (2^3 + 2^4)+...+ (2^9 + 2^{10})$$

$$A = 2(1+2) + 2^3(1+2) + ... + 2^9(1+2)$$

$$A = 2 \times 3 + 2^3 \times 3 + ... + 2^9 \times 3$$

$$A = (2 +2^3 + ... + 2^9) \times 3$$

Look! The series has x3! So A is divisible by 3.

2 years, 2 months ago

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An easier way to prove that it is divisible by $$3$$ is by using modular arithmetic. Let the given expression be $$Z$$.

$Z=\sum_{i=1}^{10} 2^i \equiv \sum_{i=1}^{10} (-1)^i\pmod{3}\\ \implies Z\equiv \left(\sum_{i=1}^5 (-1)^{2i}\right)+\left(\sum_{i=1}^{5} (-1)^{2i-1}\right)\pmod{3}\\ \implies Z\equiv \left(\sum_{i=1}^5 (1)\right)+\left(\sum_{i=1}^5 (-1)\right)\equiv \sum_{i=1}^5 (1-1)\pmod{3}\\ \implies Z\equiv \sum_{i=1}^5 (0)\pmod{3}\\ \implies Z\equiv 0\pmod{3}\\ \implies 3\mid Z$

where $$a\mid b$$ denotes that $$a$$ divides $$b$$, or in other words, $$b$$ is divisible by $$a$$. · 2 years, 1 month ago

I'm an 7th grade student, so I just use original numbers :)) Sorry, can't understand. · 2 years, 1 month ago

It's nothing much complicated. You can review the modular arithmetic wikis or ask me which part of it you can't understand?! I'd be more than glad to help you out. :) · 2 years, 1 month ago

What is the i? · 2 years, 1 month ago

$$i$$ is a counter variable in the summation. Suppose, you have the sum $$S=2^1+2^2+2^3+\ldots+2^6$$. You can write it as,

$S=\sum_{i=1}^6 2^i$

The initial and final values of $$i$$ are given in lower and upper sides of the "sigma" symbol and the value of $$i$$ gets incremented in each summation step. · 2 years, 1 month ago