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Buktikanlah. b. 3 + 3 2 + 3 3 + ... + 3 n = 2 1 ​ ( 3 n + 1 − 3 )

Buktikanlah.

b.  

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A. Acfreelance

Master Teacher

Mahasiswa/Alumni UIN Walisongo Semarang

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Pembuktian menggunakan induksi matematika Dimana untuk n = 1 Untuk n = k diasumsikan terbukti maka Untuk n = k+1 maka Jadi terbukti bahwa karena hasil dari sisi kanan dan kiri sama

Pembuktian menggunakan induksi matematika

Dimana untuk n = 1

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n end cell equals cell 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses end cell row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of 1 end cell equals cell 1 half open parentheses 3 to the power of 1 plus 1 end exponent minus 3 close parentheses end cell row 3 equals cell 1 half open parentheses 3 squared minus 3 close parentheses end cell row 3 equals cell 1 half open parentheses 6 close parentheses end cell row 3 equals cell 3 rightwards arrow Terbukti space end cell end table

Untuk n = k diasumsikan terbukti maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n end cell equals cell 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses end cell row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight k end cell equals cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses end cell row cell 3 to the power of straight k end cell equals cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses end cell row cell 3 to the power of straight k end cell equals cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses rightwards arrow Terbukti space end cell end table

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n end cell equals cell 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses end cell row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight k plus 3 to the power of straight k plus 1 end exponent end cell equals cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses end cell row cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses plus 3 to the power of straight k plus 1 end exponent end cell equals cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses end cell row cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses end cell equals cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses rightwards arrow Terbukti space end cell end table

Jadi terbukti bahwa 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n equals 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses karena hasil dari sisi kanan dan kiri sama

 

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