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Question

B= ($
3 +\frac{1}{3} )^{2} +(3^{2} +\frac{1}{3^{2}} )^{2 } +................................+(3^{n} + \frac{1}{3^{n}} )^{2}
$

score 2 · Answer 1

Bình chọn tăng 2 Bình chọn giảm

Trước hết ta có : $1+a+2+2=a^3+...+a^k=\frac{a^{k+1}-1}{a-1}$
$B=(3 +\frac{1}{3} )^{2} +(3^{2} +\frac{1}{3^{2}} )^{2 } +...+(3^{n} + \frac{1}{3^{n}} )^2=(3^2+2+\frac{1}{3^2})+(3^{2.2}+2+\frac{1}{3^{2.2}})$
$+....+(3^{2n}+2+\frac{1}{3^{2n}})=2n+(3^2+3^{2.2}+...+3^{2n})+(\frac{1}{3^2}+\frac{1}{3^{2.2}}+...+\frac{1}{3^{2n}})$
$=2n+3^2\frac{3^{2n}-1}{3^2-1}+\frac{1}{3^2}\frac{\frac{1}{3^{2n}}-1}{\frac{1}{3^2}-1}$