Pertanyaan

Nilai dari k = 3 ∑ 10 ( k + 1 ) 2 = ....

Nilai dari $k = 3 \sum 10 (k + 1)^{2} = ....$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Misal, . sehingga Ingat bahwa maka Pada induksi matematika, Jadi, Oleh karena itu, jawaban yang benar adalah B.

Misal, k plus 1 equals y rightwards arrow k equals y minus 1 . sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from k equals 3 to 10 of open parentheses k plus 1 close parentheses squared end cell equals cell sum from y minus 1 equals 3 to 10 of open parentheses y close parentheses squared end cell row blank equals cell sum from y equals 4 to 11 of open parentheses y close parentheses squared end cell end table

Ingat bahwa

maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from y equals 4 to 11 of open parentheses y close parentheses squared end cell equals cell sum from y equals 1 to 11 of open parentheses y close parentheses squared minus sum from y equals 1 to 3 of open parentheses y close parentheses squared end cell end table

Pada induksi matematika,

$table attributes columnalign right center left columnspacing 0px end attributes row cell sum from y equals 1 to n of open parentheses y close parentheses squared end cell equals cell fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction end cell row cell sum from y equals 1 to 3 of open parentheses y close parentheses squared end cell equals cell fraction numerator 3 open parentheses 3 plus 1 close parentheses open parentheses 2 cross times 3 plus 1 close parentheses over denominator 6 end fraction end cell row blank equals cell fraction numerator 3 cross times 4 cross times 7 over denominator 6 end fraction end cell row blank equals cell 14 space.................. left parenthesis 1 end cell row cell sum from y equals 1 to 11 of open parentheses y close parentheses squared end cell equals cell fraction numerator 11 open parentheses 11 plus 1 close parentheses open parentheses 2 cross times 11 plus 1 close parentheses over denominator 6 end fraction end cell row blank equals cell fraction numerator 11 cross times 12 cross times 23 over denominator 6 end fraction end cell row blank equals cell 506 space................. left parenthesis 2 end cell row blank rightwards arrow cell sum from y equals 1 to 11 of open parentheses y close parentheses squared minus sum from y equals 1 to 4 of open parentheses y close parentheses squared equals 506 minus 14 end cell row blank equals 492 end table$

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row cell sum from y equals 4 to 11 of open parentheses y close parentheses squared end cell equals 492 end table

Oleh karena itu, jawaban yang benar adalah B.