Pertanyaan

Nilai dari k = 1 ∑ 5 ( 2 k k − 1 ) adalah...

Nilai dari $k = 1 \sum 5 (\frac{k - 1}{2 k})$ adalah...

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Habis dalam

Klaim

G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.

Pembahasan

Subtitusi ke Jumlahkan semuanya, Jadi,nilai dari adalah . Oleh karena itu, jawaban yang benar adalah E.

Subtitusi k equals open curly brackets 1 comma 2 comma 3 comma 4 comma 5 close curly brackets ke $a subscript k equals fraction numerator k minus 1 over denominator 2 k end fraction$

$table attributes columnalign right center left columnspacing 0px end attributes row cell a subscript 1 end cell equals cell fraction numerator 1 minus 1 over denominator 2 times thin space 1 end fraction end cell row blank equals cell fraction numerator 0 over denominator 2 times thin space 1 end fraction end cell row blank equals cell 0 over 2 end cell row blank equals 0 row blank blank blank row cell a subscript 2 end cell equals cell fraction numerator 2 minus 1 over denominator 2 times thin space 2 end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 times thin space 2 end fraction end cell row blank equals cell 1 fourth end cell row blank blank blank row cell a subscript 3 end cell equals cell fraction numerator 3 minus 1 over denominator 2 times thin space 3 end fraction end cell row blank equals cell 2 over 6 end cell row blank equals cell 1 third end cell row blank blank blank row blank blank cell a equals presubscript 4 fraction numerator 4 minus 1 over denominator 2 times thin space 4 end fraction end cell row blank equals cell 3 over 8 end cell row blank blank blank row cell a subscript 5 end cell equals cell fraction numerator 5 minus 1 over denominator 2 times thin space 5 end fraction end cell row blank equals cell 4 over 10 end cell row blank equals cell 2 over 5 end cell end table$

Jumlahkan semuanya,

$table attributes columnalign right center left columnspacing 0px end attributes row cell 0 plus 1 fourth plus 1 third plus 3 over 8 plus 2 over 5 end cell equals cell fraction numerator 1 cross times 30 over denominator 4 cross times 30 end fraction plus fraction numerator 1 cross times 40 over denominator 3 cross times 40 end fraction plus fraction numerator 3 cross times 15 over denominator 8 cross times 15 end fraction plus fraction numerator 2 cross times 24 over denominator 5 cross times 24 end fraction end cell row blank equals cell 30 over 120 plus 40 over 120 plus 45 over 120 plus 48 over 120 end cell row blank equals cell fraction numerator 30 plus 40 plus 45 plus 48 over denominator 120 end fraction end cell row blank equals cell 163 over 120 end cell row blank equals cell 1 43 over 120 end cell end table$

Jadi, nilai dari $sum from k equals 1 to 5 of open parentheses fraction numerator k minus 1 over denominator 2 k end fraction close parentheses$ adalah 1 43 over 120 .

Oleh karena itu, jawaban yang benar adalah E.