Pertanyaan

Buktikan bahwa: b. P n ≡ 2 × 4 1 + 4 × 6 1 + ... + ( 2 n ) ( 2 n + 2 ) 1 = 4 ( n + 1 ) n

Buktikan bahwa:

b. $P_{n} \equiv \frac{1}{2 \times 4} + \frac{1}{4 \times 6} + ... + \frac{1}{( 2 n ) ( 2 n + 2 )} = \frac{n}{4 ( n + 1 )}$

A. Acfreelance

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terbukti bahwa benar karena hasil dari pembuktian ruas kanan dan kiri sama

terbukti bahwa $straight P subscript straight n identical to fraction numerator 1 over denominator 2 cross times 4 end fraction plus fraction numerator 1 over denominator 4 cross times 6 end fraction plus... plus fraction numerator 1 over denominator left parenthesis 2 straight n right parenthesis left parenthesis 2 straight n plus 2 right parenthesis end fraction equals fraction numerator straight n over denominator 4 open parentheses straight n plus 1 close parentheses end fraction$ benar karena hasil dari pembuktian ruas kanan dan kiri sama

Pembahasan

Pembuktian menggunakan induksi matematika Untuk n = 1 Untuk n = k diasumsikan benar Akan dibuktikan untuk n = k+1 Jadi terbukti bahwa benar karena hasil dari pembuktian ruas kanan dan kiri sama

Pembuktian menggunakan induksi matematika

Untuk n = 1

Untuk n = k diasumsikan benar

Akan dibuktikan untuk n = k+1

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator 2 cross times 4 end fraction plus fraction numerator 1 over denominator 4 cross times 6 end fraction plus... plus fraction numerator 1 over denominator left parenthesis 2 straight n right parenthesis left parenthesis 2 straight n plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight n over denominator 4 open parentheses straight n plus 1 close parentheses end fraction end cell row cell fraction numerator 1 over denominator 2 cross times 4 end fraction plus fraction numerator 1 over denominator 4 cross times 6 end fraction plus... plus fraction numerator 1 over denominator left parenthesis 2 straight k right parenthesis left parenthesis 2 straight k plus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis 2. straight k plus 1 right parenthesis left parenthesis 2 left parenthesis straight k plus 1 right parenthesis plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis straight k plus 1 plus 1 right parenthesis end fraction end cell row cell fraction numerator straight k over denominator 4 left parenthesis straight k plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis straight k plus 1 right parenthesis left parenthesis 2 straight k plus 2 plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis k plus 2 right parenthesis end fraction end cell row cell fraction numerator straight k over denominator 4 left parenthesis k plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis straight k plus 1 right parenthesis left parenthesis 2 straight k plus 4 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis k plus 2 right parenthesis end fraction end cell row cell fraction numerator straight k over denominator 4 left parenthesis k plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis straight k plus 1 right parenthesis 2 left parenthesis straight k plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis k plus 2 right parenthesis end fraction end cell row cell fraction numerator straight k over denominator 4 left parenthesis k plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator 4 left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis k plus 2 right parenthesis end fraction end cell row cell fraction numerator straight k left parenthesis straight k plus 2 right parenthesis over denominator 4 left parenthesis k plus 1 right parenthesis left parenthesis k plus 2 right parenthesis end fraction plus fraction numerator 1 over denominator 4 left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis k plus 2 right parenthesis end fraction end cell row cell fraction numerator k squared plus 2 k plus 1 over denominator 4 left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis k plus 2 right parenthesis end fraction end cell row cell fraction numerator up diagonal strike left parenthesis k plus 1 right parenthesis end strike left parenthesis k plus 1 right parenthesis over denominator 4 up diagonal strike left parenthesis straight k plus 1 right parenthesis end strike left parenthesis straight k plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis k plus 2 right parenthesis end fraction end cell row cell fraction numerator straight k plus 1 over denominator 4 left parenthesis straight k plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight k plus 1 over denominator 4 left parenthesis straight k plus 2 right parenthesis end fraction rightwards arrow Terbukti space end cell end table$

Jadi terbukti bahwa $straight P subscript straight n identical to fraction numerator 1 over denominator 2 cross times 4 end fraction plus fraction numerator 1 over denominator 4 cross times 6 end fraction plus... plus fraction numerator 1 over denominator left parenthesis 2 straight n right parenthesis left parenthesis 2 straight n plus 2 right parenthesis end fraction equals fraction numerator straight n over denominator 4 open parentheses straight n plus 1 close parentheses end fraction$ benar karena hasil dari pembuktian ruas kanan dan kiri sama

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