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I wanna find the sum of divisors of a number defined as:$$2^n$$ Let the series be S, $$ so, S = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 +.....2^n $$

Noww, if we factor out the 2 from 2nd term to last term, S can be written like :$$ S= 2^0 + 2(2^0 + 2^1 +2^2 + 2^{(n-1)}) $$

so. $$S = 2^0 + 2S^{(n-1)}$$

I don't know where to go from here :/

Khosrotash
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Chandler Bing
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  • Hint: remember that $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b+ a^{n-3}b^2+ \cdots + b^{n-1})$. Then let $a=2,b=1$. – dxiv Oct 16 '16 at 05:52

1 Answers1

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Hint: If you let $$S = 2^0+2^1+\cdots+2^{n-1}+2^n$$ then $$2S = 2^1+2^2+\cdots+2^n+2^{n+1}.$$

Subtract the first equation from the second to get $$2S-S = (2^1+2^2+\cdots+2^n+2^{n+1}) - (2^0+2^1+\cdots+2^{n-1}+2^n)$$

You should notice that a lot of terms cancel out. Can you figure out the formula from here?

JimmyK4542
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  • Yes, can you help me find a formula for sum of all divisors of any given integer 'n' . It would be a sum of all the combinations of factors, but how to generalize it ? – Chandler Bing Oct 18 '16 at 10:09