I have the highest regard for Giorgiana - she answered a real tough question of mine in an outstanding way.

However, I disagree with this answer.

I have both an algebraic way and a non- algebraic way to do this.

Note that each sum gets you half way between a starting point of first starting with 0 and 2

case in point 1 is 1/2 way between 0 and 2 and you have 1 to go.

add 1/2 to 1 you are half way between 1 and 2 and only have 1/2 to go.

Sum = 2 - 1/2^5 = 1 31/32

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another way:

consider sum*2 = 2 + 1 +1/2 +.....1/2^4

subtract sum = 1 + 1/2 +..............1/2^5

2 - 1/2^5 what I just said a different way.

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clearly the sum is greater than 1 since that is your first term and you add positive numbers to it.

As we can notice, we have to calculate the sum of 6 terms, which are the terms of a geometrical progression,whose first term is 1 and the ratio, q=1/2.

S6=b1*(1-q^6)/(1-q)

We've put in formula (1-q^6)/(1-q), because the ratioq=1/2<1.

S6= [1-(1/2)^6]/(1-1/2)

S6= (1-1/8)(1+1/8)/(1/2)

S6= [(2^6-1)/2^6]/1/2

S6= (2^6-1)/2^5

**S6=(64-1)/32**

**S6=63/32**