Question

Cum se rezolva?
s=1-2+2^2-2^3+...+2^100

Answer 1

S   = 1-2+2²-2³+...+2¹⁰⁰

2S =    2-2²+2³-...-2¹⁰⁰+2¹⁰¹

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S+2S = 1+2¹⁰¹

3S = 1+2¹⁰¹

=> S = (2¹⁰¹+1)/3

Answer 2

a0×(q^(n+1)-1)/(q-1)=1×(-2)^101-1)/(-2-1)=-2^101-1/(-3)

-(2^101+1)/(-3)=(1+2^101)/3=(2^101+1)/3

1-2=-1

4-8=-4=-2^2

16-32=-16=-2^4

2^98-2^99=(1-2)×2^98=-1×2^98=-2^98

=>

-(1+2^2+2^4+...+2^98)+2^100=-(1+2^2+(2^2)^2+...+(2^49)^2)+2^100=-((2^2)^50-1)/(2^2-1)+2^100=-(2^100-1)/3+2^100=(3×2^100-2^100+1)/3=(2^101+1)/3