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Misalkan Sn​ merupakan suku ke-n dari deret geometri. Buktikan bahwa:  Sn​S2n​​=1+rn

Pertanyaan

Misalkan S subscript n merupakan suku ke-n dari deret geometri. Buktikan bahwa: 

S subscript 2 n end subscript over S subscript n equals 1 plus r to the power of n 

L. Rante

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Jawaban

terbukti bahwa S subscript 2 n end subscript over S subscript n equals 1 plus r to the power of n

Pembahasan

Catatan: maksud soal S subscript n merupakan jumlah n suku pertama dari deret geometri

Ingat!

  • Jumlah n suku pertama pada deret geometri adalah

 S subscript n equals fraction numerator a open parentheses r to the power of n minus 1 close parentheses over denominator r minus 1 end fraction rightwards arrow r less than negative 1 space atau space r greater than 1 S subscript n equals fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction rightwards arrow negative 1 less than r less than 1 

  • a squared minus b squared equals open parentheses a plus b close parentheses open parentheses a minus b close parentheses 

Perhatikan perhitungan berikut!

Untuk r less than negative 1 space atau space r greater than 1 

table attributes columnalign right center left columnspacing 0px end attributes row cell S subscript 2 n end subscript over S subscript n end cell equals cell fraction numerator fraction numerator a open parentheses r to the power of 2 n end exponent minus 1 close parentheses over denominator r minus 1 end fraction over denominator fraction numerator a open parentheses r to the power of n minus 1 close parentheses over denominator r minus 1 end fraction end fraction end cell row blank equals cell fraction numerator r to the power of 2 n end exponent minus 1 over denominator r to the power of n minus 1 end fraction end cell row blank equals cell fraction numerator open parentheses r to the power of n close parentheses squared minus open parentheses 1 close parentheses squared over denominator r to the power of n minus 1 end fraction end cell row blank equals cell fraction numerator open parentheses r to the power of n plus 1 close parentheses open parentheses r to the power of n minus 1 close parentheses over denominator r to the power of n minus 1 end fraction end cell row blank equals cell r to the power of n plus 1 end cell row blank equals cell 1 plus r to the power of n end cell end table 

Untuk negative 1 less than r less than 1

 table attributes columnalign right center left columnspacing 0px end attributes row cell S subscript 2 n end subscript over S subscript n end cell equals cell fraction numerator fraction numerator a open parentheses 1 minus r to the power of 2 n end exponent close parentheses over denominator 1 minus r end fraction over denominator fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction end fraction end cell row blank equals cell fraction numerator 1 minus r to the power of 2 n end exponent over denominator 1 minus r to the power of n end fraction end cell row blank equals cell fraction numerator open parentheses 1 close parentheses squared minus open parentheses r to the power of n close parentheses squared over denominator 1 minus r to the power of n end fraction end cell row blank equals cell fraction numerator open parentheses 1 plus r to the power of n close parentheses open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r to the power of n end fraction end cell row blank equals cell 1 plus r to the power of n end cell end table 

Dengan demikian, terbukti bahwa S subscript 2 n end subscript over S subscript n equals 1 plus r to the power of n

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