s is a real number; s=1+1/2+(1/2^2)+....+(1/2^2008). Demonstrate that "s" is in the bracket (1;2).

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We've noticed that we have an addition of the terms of a geometrical progression with the ration q=1/2.

We know that the sum of n terms of a geometrical progression is written in this way:

Sn=(q^n-1)/q-1, when q>1, and Sn= (1-q^n)/1-q, q<1

We've noticed that q=1/2<1 and n=2009

S2008=s=(1-1/2^2009)/1-1/2=2-1/2^2008

1/2^2008<2, so s is in the bracket (1,2)

S=1+1/2+1/2^2+---1/2008, To show s is in[1.2]

We know that, (1- x^(n+1))/(1-x) = 1+x+x^2+x^3+...+x^n

Putting x=1/2 on the right it becomes and n=2008,

S=1+ (1/2)+(1/3)^2+...+(1/2)^2008

= {1-(1/2)^(2008+1)}/(1-1/2) = 2{1-(1/2)^2009)

=**2-1/2008**.

Therefore **S=2-1/2^2008**. So S belongs to braket [1,2].

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