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Compute the unique positive integer \(n\) such that \(2*2^2+3*2^3+4*2^4+5*2^5+...+n*2^n=2^{n+10}\)
 

 Jul 22, 2020
 #1
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a=2;b=0; c=a* 2^a;b=b+c; if(b==2^(a+10), goto5, goto6);printb, a; a++;if(a<1000, goto2, 0)

 

n = 513

 Jul 23, 2020
 #2
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sumfor(n, 0, 511,2^(n + 2)* (n + 2) = 2.745919064 E+157

 

Note: Each term of this sequence can be represented by this "closed form": 2^(n + 2)* (n + 2). I don't know how to maniplulate it algebraically to find "n", but  maybe Alan, heureka, Max, Melody and others can sum it up as geometric sequence. The first term is: (2^(n+1))*(n+1) and the common ratio:  ((2 *(n + 2))/(n + 1)) 

 Jul 23, 2020
 #3
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+2

Here's one way of doing this:

 

 Jul 23, 2020
 #4
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Thank you, all three of you. I appreciated it very much.

Firebolt  Jul 23, 2020

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