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+22)+(23+24)+...+(29+210)
A=2(1+2)+23(1+2)+...+29(1+2)
A=2×3+23×3+...+29×3
A=(2+23+...+29)×3
Look! The series has x3! So A is divisible by 3.
#Calculus
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Comments
Yes it is divisible!!!!!!!!!!!!!
An easier way to prove that it is divisible by 3 is by using modular arithmetic. Let the given expression be Z.
Z=i=1∑102i≡i=1∑10(−1)i(mod3)⟹Z≡(i=1∑5(−1)2i)+(i=1∑5(−1)2i−1)(mod3)⟹Z≡(i=1∑5(1))+(i=1∑5(−1))≡i=1∑5(1−1)(mod3)⟹Z≡i=1∑5(0)(mod3)⟹Z≡0(mod3)⟹3∣Z
where a∣b denotes that a divides b, or in other words, b is divisible by a.
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I'm an 7th grade student, so I just use original numbers :)) Sorry, can't understand.
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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. :)
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i is a counter variable in the summation. Suppose, you have the sum S=21+22+23+…+26. You can write it as,
S=i=1∑62i
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.