2024학년도 11월 미적분 29번

29. 첫째항과 공비가 각각 $0$$0$이 아닌 두 등비수열 $\{a_n\},\{b_n\}$$\{a_n\},~\{b_n\}$에 대하여 두 급수 $\sum _{n=1}^{\mathrm{\infty }}{a}_n,\sum _{n=1}^{\mathrm{\infty }}{b}_n$$\sum\limits_{n=1}^{\infty}a_n,~\sum\limits_{n=1}^{\infty}b_n$이 각각 수렴하고 $\sum _{n=1}^{\mathrm{\infty }}{a}_n{b}_n=(\sum _{n=1}^{\mathrm{\infty }}{a}_n)\times (\sum _{n=1}^{\mathrm{\infty }}{b}_n),3\times \sum _{n=1}^{\mathrm{\infty }}|a_{2n}|=7\times \sum _{n=1}^{\mathrm{\infty }}|a_{3n}|$$\sum\limits_{n=1}^{\infty} a_nb_n=\Big(\sum\limits_{n=1}^{\infty} a_n\Big)\times \Big(\sum\limits_{n=1}^{\infty} b_n\Big),~3\times \sum\limits_{n=1}^{\infty} |a_{2n}| =7\times \sum\limits_{n=1}^{\infty} |a_{3n}|$이 성립한다. $\sum _{n=1}^{\mathrm{\infty }}\frac{b_{2n-1}+b_{3n+1}}{b_n}=S$$\sum\limits_{n=1}^{\infty} \cfrac {b_{2n-1}+b_{3n+1}}{b_n}=S$일 때, $120S$$120S$의 값을 구하시오.

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i. 생각

• $a_n=a{r}_1^{n-1},b_n=b{r}_2^{n-1}$$a_n=ar_1^{n-1},~b_n=br_2^{n-1}$이라 하면, $a_n{b}_n=ab(r_1{r}_2{)}^{n-1}$$a_nb_n=ab(r_1r_2)^{n-1}$ (단, $|r_1|<1,|r_2|<1$$|r_1|<1,~|r_2|<1$)

  첫번째 조건식을 이용하자.

  $\frac{ab}{1-r_1{r}_2}=\frac{a}{1-r_1}\frac{b}{1-r_2}$$\cfrac {ab}{1-r_1r_2} =\cfrac a{1-r_1}\cfrac {b}{1-r_2}$

  정리하면, $2r_1{r}_2=r_1+r_2$$2r_1r_2=r_1+r_2$

• 두번째 조건식을 활용하자.

  i. $0<r_1<1$$0<r_1<1$

  • $a_{2n}=a{r}_1(r_1^2{)}^{n-1}$$a_{2n}=ar_1 (r_1^2 )^{n-1}$

  • $a_{3n}=a{r}_1^2(r_1^3{)}^{n-1}$$a_{3n}=ar_1^2(r_1^3)^{n-1}$

  • $3\frac{a{r}_1}{1-r_1^2}=7\frac{a{r}_1^2}{1-r_1^3}$$3\cfrac {ar_1}{1-r_1^2}=7\cfrac {ar_1^2}{1-r_1^3}$

    이 식을 정리하면,

    $4r_1^2+4r_1-3=0$$4r_1^2 +4r_1-3=0$

    $r_1=\frac{1}{2}$$r_1=\cfrac 12$

  • $2r_1{r}_2=r_1+r_2$$2r_1r_2=r_1+r_2$에 대입해 $r_2$$r_2$를 구하면,

    $r_2=\frac{1}{2}+r_2$$r_2=\cfrac 12+r_2$

    .....모순...

  ii. $-1<r<0$$-1<r<0$일 때,

  • $-\frac{a{r}_1}{1-r_1^2}=7\frac{a{r}_1^2}{1+r_1^3}$$-\cfrac {ar_1} {1-r_1^2} =7\cfrac {ar_1^2}{1+r_1^3}$

    이 식을 풀면,

    $4r_1^2-4r_1-3=0$$4r_1^2 -4r_1-3=0$

    $\therefore r_1=-\frac{1}{2}$$\therefore~ r_1=-\cfrac 12$

  • $r_2$$r_2$를 구하자.

    $-r_2=-\frac{1}{2}+r_2$$-r_2=-\cfrac 12 +r_2$

    $\therefore r_2=\frac{1}{4}$$\therefore~ r_2=\cfrac 14$

$a,b$$a,~b$를 구하지 않아도 문제가 풀리려나?

ii. 계산

• $\sum _{n=1}^{\mathrm{\infty }}\frac{b_{2n-1}+b_{3n+1}}{b_n}$$\sum\limits_{n=1}^{\infty} \cfrac {b_{2n-1}+b_{3n+1}}{b_n}$를 계산하자.

  • $b_{2n-1}=b(\frac{1}{4}{)}^{2n-1-1}=b(\frac{1}{4}{)}^{2(n-1)}=16b(\frac{1}{4}{)}^{2n}$$b_{2n-1} =b(\cfrac 14) ^{2n-1-1}=b(\cfrac 1{4})^{2(n-1)}=16b(\cfrac 14)^{2n}$

  • $b_{3n+1}=b(\frac{1}{4}{)}^{3n+1-1}=b(\frac{1}{4}{)}^{3n}$$b_{3n+1} =b(\cfrac 14)^{3n+1-1}=b(\cfrac 14)^{3n}$

  $\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{16b(\frac{1}{4}{)}^{2n}+b(\frac{1}{4}{)}^{3n}}{4b(\frac{1}{4}{)}^n}& =\sum _{n=1}^{\mathrm{\infty }}(4(\frac{1}{4}{)}^n+\frac{1}{4}(\frac{1}{4}{)}^{2n})\\ & =\frac{1}{1-\frac{1}{4}}+\frac{\frac{1}{64}}{1-\frac{1}{16}}\\ & =\frac{4}{3}+\frac{1}{60}\\ & =\frac{81}{60}\end{array}$$\begin{aligned} \sum\limits_{n=1}^{\infty} \cfrac {16b(\cfrac 14)^{2n}+b(\cfrac 14)^{3n}}{4b(\cfrac 14)^n}&=\sum\limits_{n=1}^{\infty}\Big(4(\cfrac 14)^n+\cfrac 14(\cfrac 14)^{2n}\Big) \\&=\cfrac {1}{1-\cfrac 14}+\cfrac{\cfrac 1{64}}{1-\cfrac 1{16}}\\ &=\cfrac 43+\cfrac 1{60} \\ &= \cfrac {81}{60} \end{aligned}$

$\therefore 120S=162$$\therefore~ 120S=162$