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.
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Top NewestAn 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 \(a\). – Prasun Biswas · 2 years, 3 months ago
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I'm an 7th grade student, so I just use original numbers :)) Sorry, can't understand. – Adam Phúc Nguyễn · 2 years, 3 months ago
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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. :) – Prasun Biswas · 2 years, 3 months ago
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What is the i? – Adam Phúc Nguyễn · 2 years, 3 months ago
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\(i\) 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. – Prasun Biswas · 2 years, 3 months ago
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Yes it is divisible!!!!!!!!!!!!! – Sudhir Aripirala · 2 years, 4 months ago
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