2025년 10월 미적분 29번

29. 등비수열 $\{a_n\}$$\{a_n\}$에 대하여 $\sum _{n=1}^{\mathrm{\infty }}(a_n+a_{n+1})=5,\sum _{n=1}^{\mathrm{\infty }}(|a_{n+1}+a_{n+2}|\times \sin \frac{n\pi }{2})=2$$\sum\limits_{n=1}^{\infty} \big(a_n+a_{n+1}\big)=5,~\sum\limits_{n=1}^{\infty} \Big(\big|a_{n+1}+a_{n+2}\big|\times \sin \cfrac {n\pi}2\Big)=2$일 때, $\sum _{n=1}^{\mathrm{\infty }}(100a_n-m{a}_{3n})$$\sum\limits_{n=1}^{\infty} \big(100a_n -ma_{3n} \big)$의 값이 자연수가 되도록 하는 자연수 $m$$m$의 최댓값을 구하시오.

 

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i. 정리

• $a_n=a{r}^{n-1},|r|<1$$a_n=ar^{n-1},~|r|<1$

• $\sum _{n=1}^{\mathrm{\infty }}(a_n+a_{n+1})=5$$\sum\limits_{n=1}^{\infty} \big(a_n+a_{n+1}\big)=5$를 살펴보자.

  $b_n=a_n+a_{n+1}$$b_n=a_n+a_{n+1}$이라 하면,

  $b_n=a{r}^{n-1}+a{r}^n=a{r}^{n-1}(1+r)$$b_n=ar^{n-1}+ar^n=ar^{n-1}(1+r)$

  $\therefore b_n$$\therefore~ b_n$은 첫번째항이 $a(1+r)$$a(1+r)$이고 공비는 $r$$r$인 등비수열이다.

  $b_n=a(1+r)r^{n-1}$$b_n=a(1+r)r^{n-1}$

   

  $\sum _{n=1}^{\mathrm{\infty }}(a_n+a_{n+1})=\frac{a(1+r)}{1-r}=5$$\sum\limits_{n=1}^{\infty} \big(a_n+a_{n+1}\big)=\cfrac {a(1+r)}{1-r}=5$

  $a>0$$a>0$이다.

• $\sum _{n=1}^{\mathrm{\infty }}(|a_{n+1}+a_{n+2}|\times \sin \frac{n\pi }{2})=2$$\sum\limits_{n=1}^{\infty} \Big(\big|a_{n+1}+a_{n+2}\big|\times \sin \cfrac {n\pi}2\Big)=2$를 살펴보자.

  • $\sin \frac{n\pi }{2}$$\sin\cfrac{n\pi}2$를 살펴보면,

    $1,0,-1,0,1,\cdots$$1,~0,~-1,~0,~1,~\cdots$

  • $|a_{n+1}+a_{n+2}|$$|a_{n+1}+a_{n+2}|$를 살펴보자.

    • $0<r<1$$0<r<1$이면,

      주어진 무한급수 식은,

      $(a_2+a_3)-(a_4+a_5)+(a_6+a_7)+\cdots$$(a_2+a_3)-(a_4+a_5)+(a_6+a_7)+\cdots$

      $\sum _{n=1}^{\mathrm{\infty }}(a_{2n}+a_{2n+1})(-1{)}^{n-1}=2$$\sum\limits_{n=1}^{\infty} \Big(a_{2n} +a_{2n+1}\Big)(-1)^{n-1}=2$

      $\sum _{n=1}^{\mathrm{\infty }}ar(1+r)(-r^2{)}^{n-1}=2$$\sum\limits_{n=1}^{\infty} ar(1+r)(-r^2)^{n-1}=2$

       

      $\frac{ar(1+r)}{1+r^2}=2$$\cfrac {ar(1+r)}{1+r^2}=2$

       

      두 식을 연립하면,

      $\frac{\frac{a(1+r)}{1-r}}{\frac{ar(1+r)}{1+r^2}}=\frac{1+r^2}{r(1-r)}=\frac{5}{2}$$\cfrac{\cfrac {a(1+r)}{1-r}}{\cfrac {ar(1+r)}{1+r^2}}=\cfrac {1+r^2}{r(1-r)}=\cfrac 52$

       

      $2+2r^2=5r-5r^2⟶7r^2-5r+2=0$$2+2r^2 =5r-5r^2\quad \longrightarrow 7r^2-5r+2=0$

      $D=5^2-4\cdot 7\cdot 2=25-56=-31$$D=5^2 -4\cdot 7\cdot 2=25-56=-31$

      실근이 존재하지 않는다.....

    • $-1<r<0$$-1<r<0$이면,

      $\sum _{n=1}^{\mathrm{\infty }}(a_{2n}+a_{2n+1})(-1{)}^n=2$$\sum\limits_{n=1}^{\infty} (a_{2n}+a_{2n+1})(-1)^n=2$

      계산하면,

      $-\sum _{n=1}^{\mathrm{\infty }}(ar(1+r)(-r^2{)}^{n-1}=2$$-\sum\limits_{n=1}^{\infty} \big(ar(1+r)(-r^2)^{n-1}=2$

       

      부호만 바뀐 식이니까 $-\frac{1+r^2}{r(1-r)}=\frac{5}{2}$$-\cfrac {1+r^2}{r(1-r)}=\cfrac 52$이다.

      $-2-2r^2=5r-5r^2$$-2-2r^2=5r -5r^2$

      $3r^2-5r^2-2=0$$3r^2-5r^2-2=0$

      $(3r+1)(r-2)=0$$(3r+1)(r-2)=0$

      $\therefore r=-\frac{1}{3}$$\therefore~ r=-\cfrac 13$

    • $a$$a$를 구하면,

      $\frac{a(1+r)}{1-r}=5$$\cfrac {a(1+r)}{1-r}=5$에 대입하면,

      $a\frac{\frac{2}{3}}{\frac{4}{3}}=5$$a\cfrac {\cfrac 23}{\cfrac 43}=5$

      $a=10$$a=10$

$\therefore a_n=10(-\frac{1}{3}{)}^{n-1}$$\therefore~ a_n=10\Big(-\cfrac 13 \Big)^{n-1}$

ii. 계산...

• $\sum _{n=1}^{\mathrm{\infty }}100a_n$$\sum\limits_{n=1}^{\infty} 100a_n$를 계산하자...

  $\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}100a_n& =100\cdot \frac{10}{1+\frac{1}{3}}\\ & =50\cdot 5\cdot 3\\ & =750\end{array}$$\begin{aligned} \sum\limits_{ n=1}^{\infty} 100a_n&=100 \cdot \cfrac {10}{1+\cfrac 13}\\ &=50\cdot 5\cdot 3\\& =750\end{aligned}$

• $\sum _{n=1}^{\mathrm{\infty }}{a}_{3n}$$\sum\limits_{ n=1}^{\infty}a_{3n}$을 계산하자.

  $\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}{a}_{3n}& =\frac{\frac{10}{9}}{1+\frac{1}{27}}\\ & =\frac{\frac{10}{9}}{\frac{28}{27}}\\ & =\frac{15}{14}\end{array}$$\begin{aligned}\sum\limits_{ n=1}^{\infty}a_{3n}&= \cfrac {\cfrac {10}9}{1+\cfrac 1{27}}\\ &=\cfrac {\cfrac {10}9}{\cfrac{28}{27}} \\ &=\cfrac {15}{14}\end{aligned}$

iii. 마지막 계산!

$750-\frac{15}{14}m$$750-\cfrac {15}{14}m$가 자연수가 되는 최대의 $m$$m$값은?

$m=14k$$m=14k$라 하면, $k$$k$는 자연수

$750-15k>0$$750-15k>0$을 풀면,

$50>k$$50>k$

$\therefore k=49$$\therefore~ k=49$

$\therefore m=14\cdot 49=686$$\therefore~ m=14\cdot 49= 686$