Jawaban:

Pembahasan Soal:

Jika begin mathsize 14px style f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 minus x end fraction end style, maka nilai dari

begin mathsize 14px style f left parenthesis 2.016 right parenthesis plus f left parenthesis 2.015 right parenthesis plus... plus f left parenthesis 3 right parenthesis plus f left parenthesis 2 right parenthesis plus end style

begin mathsize 14px style f open parentheses 1 half close parentheses plus f open parentheses 1 third close parentheses plus... plus f open parentheses fraction numerator 1 over denominator 2.015 end fraction close parentheses plus f open parentheses fraction numerator 1 over denominator 2.016 end fraction close parentheses end style

begin mathsize 14px style equals open parentheses negative fraction numerator 2.016 over denominator 2.015 end fraction close parentheses plus open parentheses negative 2015 over 2014 close parentheses plus... open parentheses negative 3 over 2 close parentheses plus open parentheses negative 2 over 1 close parentheses plus space space space space open parentheses 1 over 1 close parentheses plus open parentheses 1 half close parentheses plus... open parentheses 1 over 2014 close parentheses plus open parentheses 1 over 2015 close parentheses end style

Perhatikan bahwa

begin mathsize 14px style f open parentheses 2016 close parentheses plus f open parentheses 1 over 2016 close parentheses equals negative 2016 over 2015 plus 1 over 2015 equals fraction numerator negative 2015 over denominator 2015 end fraction equals negative 1 end style

begin mathsize 14px style f open parentheses 2015 close parentheses plus f open parentheses 1 over 2015 close parentheses equals fraction numerator negative 2015 over denominator 2014 end fraction plus 1 over 2014 equals fraction numerator negative 2014 over denominator 2014 end fraction equals negative 1 end style

begin mathsize 14px style f open parentheses 3 close parentheses plus f open parentheses 1 third close parentheses equals negative 3 over 2 plus 1 half equals negative 1 end style

begin mathsize 14px style f open parentheses 2 close parentheses plus f open parentheses 1 half close parentheses equals fraction numerator negative 2 over denominator 1 end fraction plus 1 over 1 equals fraction numerator negative 1 over denominator 1 end fraction equals negative 1 end style

ketika dijumlahkan nilai fungsi pertama dengan nilai fungsi terakhir, nilai fungsi kedua dan nilai fungsi kedua terakhir sampai nilai fungsi begin mathsize 14px style f open parentheses 2 close parentheses plus f open parentheses 1 half close parentheses end style, nilai nya sama yaitu begin mathsize 14px style negative 1 end style. maka dari begin mathsize 14px style f open parentheses 2016 close parentheses end style sampai begin mathsize 14px style f open parentheses 2 close parentheses end style terdapat begin mathsize 14px style 2015 end style deret, sehingga

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begin mathsize 14px style equals 2015 cross times open parentheses negative 1 close parentheses equals negative 2015 end style

Jadi, Tidak ada jawaban yang tep