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Sagot :
Let [tex]S_n[/tex] be the [tex]n[/tex]-th term in this sequence, so
[tex]S_1 = 1[/tex]
[tex]S_2 = 1 + 2[/tex]
[tex]S_3 = 1 + 2 + 2^2[/tex]
and so on, with general [tex]n[/tex]-th term
[tex]S_n = 1 + 2 + 2^2 + \cdots + 2^{n-1} = \displaystyle \sum_{i=1}^n 2^{i-1}[/tex]
Observe that
[tex]2 S_n = 2 + 2^2 + 2^3 + \cdots + 2^n[/tex]
and by subtracting this from [tex]S_n[/tex], we eliminate all but the outermost terms, namely
[tex]S_n - 2S_n = 1 - 2^n[/tex]
so that
[tex]-S_n = 1 - 2^n[/tex]
[tex]S_n = 2^n - 1[/tex]
Now, the sum of the first [tex]n[/tex] terms of the sequence [tex]\{S_1,S_2,S_3,\ldots,S_n\}[/tex] is
[tex]S = S_1 + S_2 + S_3 + \cdots + S_n[/tex]
[tex]S = \displaystyle \sum_{i=1}^n S_i[/tex]
[tex]S = \displaystyle \sum_{i=1}^n (2^i - 1)[/tex]
[tex]S = \displaystyle \sum_{i=1}^n 2^i - \sum_{i=1}^n 1[/tex]
[tex]S = \displaystyle 2 \underbrace{\sum_{i=1}^n 2^{i-1}}_{S_n} - n[/tex]
[tex]S = 2(2^n - 1) - n[/tex]
[tex]S = \boxed{2^{n+1} - n - 2}[/tex]
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