Đáp án:
Giải thích các bước giải:
Ta có A = $2^{0}$ + $2^{1}$ + $2^{2}$ + $2^{3}$ +...+ $2^{2006}$
=> 2A = 2.($2^{0}$ + $2^{1}$ + $2^{2}$ + $2^{3}$ +...+ $2^{2006}$)
=> 2A = $2^{1}$ + $2^{2}$ + $2^{3}$ + $2^{4}$ +...+ $2^{2007}$
=> 2A - A = ($2^{1}$ + $2^{2}$ + $2^{3}$ + $2^{4}$ +...+ $2^{2007}$) - ($2^{0}$ + $2^{1}$ + $2^{2}$ + $2^{3}$ +...+ $2^{2006}$)
=> A = $2^{2007}$ - $2^{0}$
=> A = $2^{2007}$ - 1
Mà B = $2^{2017}$ (đề cho)
Nên A = B - 1
Vậy A = B - 1