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Buktikan dengan prinsip induksi matematika.  a. k=1∑n​k(4k2−3)=2(n2+n)(2n2+2n−3)​

Pertanyaan

Buktikan dengan prinsip induksi matematika. 

a. begin mathsize 14px style sum from k equals 1 to n of k left parenthesis 4 k squared minus 3 right parenthesis equals fraction numerator left parenthesis n squared plus n right parenthesis left parenthesis 2 n squared plus 2 n minus 3 right parenthesis over denominator 2 end fraction end style  

Pembahasan:

Akan kita buktikan bahwa: 

sum from k equals 1 to n of k left parenthesis 4 k squared minus 3 right parenthesis equals 1 plus 26 plus 99 plus... plus n left parenthesis 4 n squared minus 3 right parenthesis

DIketahui:

Langkah 1. 

untuk n equals 1 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from k equals 1 to n of k left parenthesis 4 k squared minus 3 right parenthesis end cell equals cell fraction numerator left parenthesis n squared plus n right parenthesis left parenthesis 2 n squared plus 2 n minus 3 right parenthesis over denominator 2 end fraction end cell row cell 1 left parenthesis 4 cross times 1 squared minus 3 right parenthesis end cell equals cell fraction numerator left parenthesis 1 squared plus 1 right parenthesis left parenthesis 2 cross times 1 squared plus 2 cross times 1 minus 3 right parenthesis over denominator 2 end fraction end cell row cell 1 left parenthesis 4 cross times 1 minus 3 right parenthesis end cell equals cell fraction numerator left parenthesis 1 plus 1 right parenthesis left parenthesis 2 cross times 1 plus 2 minus 3 right parenthesis over denominator 2 end fraction end cell row cell 1 left parenthesis 4 minus 3 right parenthesis end cell equals cell fraction numerator 2 left parenthesis 2 plus 2 minus 3 right parenthesis over denominator 2 end fraction end cell row cell 4 minus 3 end cell equals cell 4 minus 3 end cell row 1 equals cell 1 space text (benar) end text end cell end table

Langkah 2. 

untuk n equals k maka:

1 plus 26 plus 99 plus... plus k left parenthesis 4 k squared minus 3 right parenthesis equals fraction numerator left parenthesis k squared plus k right parenthesis left parenthesis 2 k squared plus 2 k minus 3 right parenthesis over denominator 2 end fraction space text (asumsi: benar) end text

Langkah 3. 

untuk n equals k plus 1 maka:

1 plus 26 plus 99 plus... plus k left parenthesis 4 k squared minus 3 right parenthesis plus left parenthesis k plus 1 right parenthesis left parenthesis 4 left parenthesis k plus 1 right parenthesis squared minus 3 right parenthesis equals fraction numerator left parenthesis k squared plus k right parenthesis left parenthesis 2 k squared plus 2 k minus 3 right parenthesis over denominator 2 end fraction plus left parenthesis k plus 1 right parenthesis left parenthesis 4 left parenthesis k plus 1 right parenthesis squared minus 3 right parenthesis equals fraction numerator left parenthesis k squared plus k right parenthesis left parenthesis 2 k squared plus 2 k minus 3 right parenthesis plus 2 left parenthesis k plus 1 right parenthesis left parenthesis 4 left parenthesis k plus 1 right parenthesis squared minus 3 right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis k left parenthesis 2 k squared plus 2 k minus 3 right parenthesis plus 2 left parenthesis 4 left parenthesis k plus 1 right parenthesis squared minus 3 right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis 2 k cubed plus 2 k squared minus 3 k plus 2 left parenthesis 4 left parenthesis k squared plus 2 k plus 1 right parenthesis minus 3 right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis 2 k cubed plus 2 k squared minus 3 k plus 8 k squared plus 16 k plus 8 minus 6 right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis 2 k cubed plus 10 k squared plus 13 k plus 2 right parenthesis over denominator 2 end fraction equals fraction numerator 2 k to the power of 4 plus 10 k cubed plus 13 k squared plus 2 k plus 2 k cubed plus 10 k squared plus 13 k plus 2 over denominator 2 end fraction equals fraction numerator 2 k to the power of 4 plus 12 k cubed plus 23 k squared plus 15 k plus 2 over denominator 2 end fraction equals fraction numerator left parenthesis k squared plus 3 k plus 2 right parenthesis left parenthesis 2 k squared plus 6 k plus 1 right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis left parenthesis k plus 1 right parenthesis squared plus left parenthesis k plus 1 right parenthesis right parenthesis left parenthesis 2 left parenthesis k plus 1 right parenthesis squared plus 2 left parenthesis k plus 1 right parenthesis minus 3 right parenthesis over denominator 2 end fraction text (benar) end text

Jadi, terbukti bahwa begin mathsize 14px style sum from k equals 1 to n of k left parenthesis 4 k squared minus 3 right parenthesis equals fraction numerator left parenthesis n squared plus n right parenthesis left parenthesis 2 n squared plus 2 n minus 3 right parenthesis over denominator 2 end fraction end style

Jawaban terverifikasi

Dijawab oleh:

R. Bella

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 07 Oktober 2021

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