Tính tổng: \(S = {\left( {2 + \frac{1}{2}} \right)^2} + {\left( {4 + \frac{1}{4}} \right)^2} + ... + {\left( {{2^n} + \frac{1}{{{2^n}}}} \right)^2}\)
A.\(2n + \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
B.\(2n - \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
C.\( - 2n + \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
D.\( - 2n - \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
A.\(2n + \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
B.\(2n - \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
C.\( - 2n + \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
D.\( - 2n - \frac{{\left( {{4^n} - 1} \right)\left( {{4^{n + 1}} + 1} \right)}}{{{{3.4}^n}}}\)
Loga
Hóa Học lớp 12
