QUIZZZ
(back to school)
Tentukan hasil
[tex]\boxed{\sf\frac{1}{3^4}+\frac{3}{5^4}+\frac{6}{7^4}+\frac{10}{9^4}+\cdots=\cdots}[/tex]
[tex]A.\frac{\pi ^2}{64}\left(1-\frac{\pi ^2}{12}\right)[/tex]
[tex]B.\frac{\pi ^2}{64}[/tex]
[tex]C.3\frac{\pi ^2}{64}[/tex]
[tex]D.\frac{\pi ^2}{64}\left(1-\frac{\pi ^2}{48}\right)[/tex]
(back to school)
Tentukan hasil
[tex]\boxed{\sf\frac{1}{3^4}+\frac{3}{5^4}+\frac{6}{7^4}+\frac{10}{9^4}+\cdots=\cdots}[/tex]
[tex]A.\frac{\pi ^2}{64}\left(1-\frac{\pi ^2}{12}\right)[/tex]
[tex]B.\frac{\pi ^2}{64}[/tex]
[tex]C.3\frac{\pi ^2}{64}[/tex]
[tex]D.\frac{\pi ^2}{64}\left(1-\frac{\pi ^2}{48}\right)[/tex]
Jawab:
D. [tex]\displaystyle \bold{\dfrac{\pi^{2} }{64}(1 -\dfrac{\pi^{2} }{12})}[/tex]
Penjelasan dengan langkah-langkah:
[tex]\sf\dfrac{1}{3^4}+\dfrac{3}{5^4}+\dfrac{6}{7^4}+\dfrac{10}{9^4}+\cdots=\cdots[/tex]
rumus suku ke-n dari barisan diatas:
[tex]\displaystyle \dfrac{\dfrac{n(n+1)}{2} }{(2n+1)^{4}}[/tex]
Jadi:
[tex]\displaystyle \sum\limits_{n=1}^{\infty}\dfrac{\dfrac{n(n+1)}{2} }{(2n+1)^{4}}[/tex]
= [tex]\displaystyle \dfrac{1}{2} \sum\limits_{n=1}^{\infty}\dfrac{n(n+1)}{(2n+1)^{4}}[/tex]
= [tex]\displaystyle \dfrac{1}{2} \sum\limits_{n=1}^{\infty}\dfrac{\dfrac{1}{4}(2n+1)^{2}-\dfrac{1}{4}}{(2n+1)^{4}}[/tex]
= [tex]\displaystyle \dfrac{1}{8} \sum\limits_{n=1}^{\infty}\dfrac{1}{(2n+1)^{2}}- \dfrac{1}{8} \sum\limits_{n=1}^{\infty}\dfrac{1}{(2n+1)^{4}}[/tex]
= [tex]\displaystyle \dfrac{1}{8} (\dfrac{\pi^{2} }{8} -1)-\frac{1}{8}\:[\frac{15}{16} .S (4) - 1][/tex]
= [tex]\displaystyle \dfrac{\pi^{2} }{64} -\frac{1}{8}.\frac{15}{16}.\dfrac{\pi^{4} }{90}[/tex]
= [tex]\displaystyle \dfrac{\pi^{2} }{64} -\dfrac{\pi^{4} }{768}[/tex]
= [tex]\displaystyle \bold{\dfrac{\pi^{2} }{64}(1 -\dfrac{\pi^{2} }{12})}[/tex]