Pertanyaan

Jumlah deret tak hingga ( 2 1 ) l o g ( 2 ) 1 + ( 4 1 ) l o g ( 2 ) 1 + ( 8 1 ) l o g ( 2 ) 1 + . . . = ....

Jumlah deret tak hingga

$(\frac{1}{2})^{\frac{1}{l o g ( 2 )}} + (\frac{1}{4})^{\frac{1}{l o g ( 2 )}} + (\frac{1}{8})^{\frac{1}{l o g ( 2 )}} + . . . =$ ....

N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D

Pembahasan

Jadi, jawaban yang tepat adalah D

$begin mathsize 14px style open parentheses 1 half close parentheses to the power of fraction numerator 1 over denominator log open parentheses 2 close parentheses end fraction end exponent plus open parentheses 1 fourth close parentheses to the power of fraction numerator 1 over denominator log open parentheses 2 close parentheses end fraction end exponent plus open parentheses 1 over 8 close parentheses to the power of fraction numerator 1 over denominator log open parentheses 2 close parentheses end fraction end exponent plus. space. space. equals end style$

$begin mathsize 14px style left parenthesis 1 half right parenthesis to the power of fraction numerator 1 over denominator log left parenthesis 2 right parenthesis end fraction end exponent plus left parenthesis 1 fourth right parenthesis to the power of fraction numerator 1 over denominator log left parenthesis 2 right parenthesis end fraction end exponent plus left parenthesis fraction numerator 1 over denominator 8 end fraction right parenthesis to the power of fraction numerator 1 over denominator log left parenthesis 2 right parenthesis end fraction end exponent plus... equals open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of fraction numerator 1 over denominator log open parentheses 2 close parentheses end fraction end exponent plus open parentheses 2 to the power of negative 2 end exponent close parentheses to the power of fraction numerator 1 over denominator log open parentheses 2 close parentheses end fraction end exponent plus open parentheses 2 to the power of negative 3 end exponent close parentheses to the power of fraction numerator 1 over denominator log left parenthesis 2 right parenthesis end fraction end exponent plus... equals open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of log presuperscript 2 space 10 end exponent plus open parentheses 2 to the power of negative 2 end exponent close parentheses to the power of log presuperscript 2 space 10 end exponent plus negative 3 to the power of blank open parentheses 2 to the power of negative 3 end exponent close parentheses to the power of log presuperscript 2 space 10 end exponent plus... equals left parenthesis 10 right parenthesis to the power of negative 1 end exponent plus open parentheses 10 close parentheses to the power of negative 2 end exponent plus open parentheses 10 close parentheses to the power of negative 3 end exponent plus... equals a equals open parentheses 10 close parentheses to the power of negative 1 end exponent r equals open parentheses 10 close parentheses to the power of negative 1 end exponent S subscript infinity equals fraction numerator a over denominator 1 minus r end fraction equals fraction numerator left parenthesis 10 right parenthesis to the power of negative 1 end exponent over denominator 1 minus open parentheses 10 close parentheses to the power of negative 1 end exponent end fraction equals fraction numerator begin display style 1 over 10 end style over denominator 1 minus begin display style 1 over 10 end style end fraction equals fraction numerator begin display style 1 over 10 end style over denominator begin display style 9 over 10 end style end fraction equals 1 over 9 end style$

Jadi, jawaban yang tepat adalah D