Pertanyaan

Buktikan masing-masing notasi sigma berikut. c. k = 1 ∑ n ( 5 k − 1 ) = 2 n ( 5 n + 3 )

Buktikan masing-masing notasi sigma berikut.

c. $k = 1 \sum n (5 k - 1) = \frac{n ( 5 n + 3 )}{2}$

A. Acfreelance

Master Teacher

Mahasiswa/Alumni UIN Walisongo Semarang

Jawaban terverifikasi

Pembahasan

Dibuktikan dengan menggunkan induksi matematika Untuk n = 1 Untuk n = k diasumsikan bahwa terbukti maka Untuk n = k+1 maka PRK Jadi menggunakan induksi matematika terbukti karena hasil dari ruas kanan dan kiri memiliki hasil yang sama

Dibuktikan dengan menggunkan induksi matematika

Untuk n = 1

$table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 1 to straight n of open parentheses 5 straight k minus 1 close parentheses end cell equals cell fraction numerator straight n open parentheses 5 straight n plus 3 close parentheses over denominator 2 end fraction end cell row cell sum from straight k equals 1 to 1 of open parentheses 5.1 minus 1 close parentheses end cell equals cell fraction numerator 1 open parentheses 5.1 plus 3 close parentheses over denominator 2 end fraction end cell row 4 equals cell 4 rightwards arrow terbukti end cell row blank blank blank end table$

Untuk n = k diasumsikan bahwa terbukti maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 1 to straight n of open parentheses 5 straight k minus 1 close parentheses end cell equals cell fraction numerator straight n open parentheses 5 straight n plus 3 close parentheses over denominator 2 end fraction end cell row cell sum from straight k equals 1 to straight k of open parentheses 5 straight k minus 1 close parentheses end cell equals cell fraction numerator straight k open parentheses 5 straight k plus 3 close parentheses over denominator 2 end fraction end cell row blank equals cell fraction numerator straight k open parentheses 5 straight k plus 3 close parentheses over denominator 2 end fraction rightwards arrow terbukti end cell row blank blank blank end table$

Untuk n = k+1 maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 1 to straight n of open parentheses 5 straight k minus 1 close parentheses end cell equals cell fraction numerator straight n open parentheses 5 straight n plus 3 close parentheses over denominator 2 end fraction end cell row cell sum from straight k equals 1 to straight k plus 1 of open parentheses 5 straight k minus 1 close parentheses end cell equals cell fraction numerator straight k plus 1 open parentheses 5 left parenthesis straight k plus 1 right parenthesis right parenthesis plus 3 close parentheses over denominator 2 end fraction end cell row blank equals cell fraction numerator left parenthesis straight k plus 1 right parenthesis open parentheses 5 k plus 8 close parentheses over denominator 2 end fraction rightwards arrow terbukti end cell row blank blank blank end table$

PRK

$table attributes columnalign right center left columnspacing 0px end attributes row blank equals cell fraction numerator straight n open parentheses 5 straight n plus 3 close parentheses over denominator 2 end fraction end cell row cell sum from straight k equals 1 to straight k plus 1 of open parentheses 5 straight k minus 1 close parentheses end cell equals cell sum from straight k equals 1 to straight k of open parentheses 5 straight k minus 1 close parentheses plus sum from straight k equals straight k plus 1 to straight k plus 1 of open parentheses 5 straight k minus 1 close parentheses end cell row cell fraction numerator open parentheses straight k plus 1 close parentheses open parentheses 5 straight k plus 8 close parentheses over denominator 2 end fraction end cell equals cell fraction numerator straight k open parentheses 5 straight k plus 3 close parentheses over denominator 2 end fraction plus open parentheses 5 open parentheses straight k plus 1 close parentheses minus 1 close parentheses end cell row cell fraction numerator open parentheses straight k plus 1 close parentheses open parentheses 5 straight k plus 8 close parentheses over denominator 2 end fraction end cell equals cell fraction numerator straight k open parentheses 5 straight k plus 3 close parentheses over denominator 2 end fraction plus open parentheses 5 straight k plus 4 close parentheses end cell row cell fraction numerator open parentheses straight k plus 1 close parentheses open parentheses 5 straight k plus 8 close parentheses over denominator 2 end fraction end cell equals cell fraction numerator 5 straight k squared plus 3 straight k plus 10 straight k plus 8 over denominator 2 end fraction end cell row cell fraction numerator open parentheses straight k plus 1 close parentheses open parentheses 5 straight k plus 8 close parentheses over denominator 2 end fraction end cell equals cell fraction numerator 5 straight k squared plus 13 straight k plus 8 over denominator 2 end fraction end cell row cell fraction numerator open parentheses straight k plus 1 close parentheses open parentheses 5 straight k plus 8 close parentheses over denominator 2 end fraction end cell equals cell fraction numerator open parentheses straight k plus 1 close parentheses open parentheses 5 straight k plus 8 close parentheses over denominator 2 end fraction rightwards arrow terbukti end cell row blank blank blank end table$

Jadi $sum from straight k equals 1 to straight n of open parentheses 5 straight k minus 1 close parentheses equals fraction numerator straight n open parentheses 5 straight n plus 3 close parentheses over denominator 2 end fraction$ menggunakan induksi matematika terbukti karena hasil dari ruas kanan dan kiri memiliki hasil yang sama

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