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Given the sequence $1 ; 2 ; 3 ; ... ; 2000$. Calculate the sum of the terms that remain when the powers of 2 are removed from the given sequence.
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powers of 2= 2^2 ; 2^3 ; 2^4; ....this is a geometric seq with ratio= 2^4/2^3 = 2^3/2^2 = 2

a=4

1;2;3;.....;2000 is an arithmetic seq with common difference = 3-2=2-1=1

a=1

arithmetic sum= n(2a +(n-1))/2 = 2000(2x1 + (2000-1))/2

= 2 001 000

sum of geometric = a(r^n -1)/(r-1) =

n= last term with value 2^10 = 1 024  , as 2^11 gives  2 048 which is out of range.

we neeed to find position of 1 024, Tn=ar^(n-1)

1 024= 4x2^(n-1)

1 024/4 = 2^(n-1)

log256= (n-1)log2

n= log256/log2 + 1 = 8+1=9

hence sum= 4(2^9 -1)/(2-1) = 2 044

Therefore sum remaining when powers of 2 removed= 2 001 000-2 044= 1 998 956
by Diamond (44,680 points)

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