자연계열 1번

(1) ① $\displaystyle b_n=b+(n-1)d$이고② $\displaystyle a_nb_n=\sum_{k=1}^n a_kb_k-\sum_{k=1}^{n-1}a_kb_k$ $\displaystyle (n\ge2)$이므로③ $\displaystyle a_nb_n=n(n+1)(2n+1)-n(n-1)(2n-1)=6n^2$ $\displaystyle (n\ge2)$④ $\displaystyle b_n\ne0$이므로 양변을 $\displaystyle b_n$으로 나누면 $\displaystyle a_n=\frac{6n^2}{b+(n-1)d}$ $\displaystyle (n\ge2)$⑤ $\displaystyle a_1b_1=1\times2\times3=6$, $\displaystyle a_1=\frac6b$이므로 $\displaystyle a_n=\frac{6n^2}{b+(n-1)d}$ $\displaystyle (n\ge1)$

(2) ① $\displaystyle a_n=\frac{6n^2}{b+(n-1)d}$, $\displaystyle b_n=b+(n-1)d$이고② $\displaystyle b_n\ne0$이므로 양변을 $\displaystyle b_n$으로 나누면③ $\displaystyle \frac{a_n}{b_n}=\frac{6n^2}{\{b+(n-1)d\}^2}=\frac6{\left\{\frac bn+\left(1-\frac1n\right)d\right\}^2}$④ $\displaystyle n\to\infty$일 때 $\displaystyle \frac1n\to0$이므로⑤ $\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac6{\left\{\frac bn+\left(1-\frac1n\right)d\right\}^2}=\frac6{d^2}$.

(3) ① $\displaystyle a_n=\frac{6n^2}{b+(n-1)d}$에서 $\displaystyle b=d$이므로 $\displaystyle a_n=\frac{6n^2}{nd}=\frac{6n}d$② $\displaystyle \sqrt{a_{k+1}}-\sqrt{a_k}\ne0$이므로③ $\displaystyle \sum_{k=1}^{10}\frac1{\sqrt{a_{k+1}}+\sqrt{a_k}}=\sum_{k=1}^{10}\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k}$④ $\displaystyle a_{k+1}-a_k=\frac6d$이므로⑤ $\displaystyle \frac d6\sum_{k=1}^{10}(\sqrt{a_{k+1}}-\sqrt{a_k})=\frac d6(\sqrt{a_{11}}-\sqrt{a_1})=\sqrt{\frac d6}(\sqrt{11}-1)=\frac{\sqrt{6d}(\sqrt{11}-1)}6$