Roboguru

Diketahui:   Carilah  c.

Pertanyaan

Diketahui:

u subscript 1 equals 4 semicolon space u subscript 2 equals 6 semicolon space u subscript 3 equals 8 semicolon space u subscript 4 equals 10 semicolon space text dan  end text u subscript 5 equals 12 v subscript 1 equals 1 semicolon space v subscript 2 equals 2 semicolon space v subscript 3 equals 4 semicolon space v subscript 4 equals 5 semicolon space text dan  end text v subscript 5 equals 6 

Carilah 

c. sum from i equals 1 to 5 of u subscript i times v subscript i 

Pembahasan Soal:

Ingat kembali mengenai definisi dan sifat notasi sigma:

Jika m less or equal than n semicolon space a subscript i comma space b subscript i space text bilangan real; end text space m comma space n space text bilangan bulat end text, maka

  • sum from i equals m to n of a subscript i equals a subscript m plus a subscript m minus 1 end subscript plus a subscript m minus 2 end subscript plus midline horizontal ellipsis plus a subscript n
  • sum from i equals m to n of a subscript i times b subscript i equals sum from i equals m to n of a subscript i times sum from i equals m to n of b subscript i

Oleh karena itu, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 1 to 5 of u subscript i times v subscript i end cell equals cell sum from i equals 1 to 5 of u subscript i times sum from i equals 1 to 5 of v subscript i end cell row blank equals cell open parentheses u subscript 1 plus u subscript 2 plus u subscript 3 plus u subscript 4 plus u subscript 5 close parentheses times open parentheses v subscript 1 plus v subscript 2 plus v subscript 3 plus v subscript 4 plus v subscript 5 close parentheses end cell row blank equals cell open parentheses 4 plus 6 plus 8 plus 10 plus 12 close parentheses times open parentheses 1 plus 2 plus 4 plus 5 plus 6 close parentheses end cell row blank equals cell 40 times 18 end cell row blank equals 720 end table   

 Dengan demikian, nilai dari sum from i equals 1 to 5 of open parentheses u subscript i minus v subscript i close parentheses adalah 720.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Ridha

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 14 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Tentukan nilai penjumlahan dari notasi sigma berikut. b.

Pembahasan Soal:

Bentuk begin mathsize 14px style sum from p equals 0 to 5 of open parentheses fraction numerator 1 over denominator p plus 1 end fraction plus fraction numerator 1 over denominator p plus 2 end fraction close parentheses end style adalah bentuk sigma yang memiliki batas bawah 0 dan batas atas 5, artinya kita akan membuat deret, dengan mensubstitusikan nilai berurut begin mathsize 14px style p end style dari 0 hingga 5 yaitu begin mathsize 14px style p equals 0 end stylebegin mathsize 14px style p equals 1 end stylebegin mathsize 14px style p equals 2 end stylebegin mathsize 14px style p equals 3 end stylebegin mathsize 14px style p equals 4 end style dan begin mathsize 14px style p equals 5 end style, maka: 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from p equals 0 to 5 of open parentheses fraction numerator 1 over denominator p plus 1 end fraction plus fraction numerator 1 over denominator p plus 2 end fraction close parentheses end cell equals cell sum from p equals 0 to 5 of open parentheses fraction numerator p plus 2 plus p plus 1 over denominator left parenthesis p plus 1 right parenthesis left parenthesis p plus 2 right parenthesis end fraction close parentheses end cell row blank equals cell sum from p equals 0 to 5 of open parentheses fraction numerator 2 p plus 3 over denominator left parenthesis p plus 1 right parenthesis left parenthesis p plus 2 right parenthesis end fraction close parentheses end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from p equals 0 to 5 of open parentheses fraction numerator 2 p plus 3 over denominator left parenthesis p plus 1 right parenthesis left parenthesis p plus 2 right parenthesis end fraction close parentheses end cell equals cell fraction numerator 2 left parenthesis 0 right parenthesis plus 3 over denominator left parenthesis 0 plus 1 right parenthesis left parenthesis 0 plus 2 right parenthesis end fraction plus fraction numerator 2 left parenthesis 1 right parenthesis plus 3 over denominator left parenthesis 1 plus 1 right parenthesis left parenthesis 1 plus 2 right parenthesis end fraction end cell row blank blank cell plus fraction numerator 2 left parenthesis 2 right parenthesis plus 3 over denominator left parenthesis 2 plus 1 right parenthesis left parenthesis 2 plus 2 right parenthesis end fraction plus fraction numerator 2 left parenthesis 3 right parenthesis plus 3 over denominator left parenthesis 3 plus 1 right parenthesis left parenthesis 3 plus 2 right parenthesis end fraction end cell row blank blank cell plus fraction numerator 2 left parenthesis 4 right parenthesis plus 3 over denominator left parenthesis 4 plus 1 right parenthesis left parenthesis 4 plus 2 right parenthesis end fraction plus fraction numerator 2 left parenthesis 5 right parenthesis plus 3 over denominator left parenthesis 5 plus 1 right parenthesis left parenthesis 5 plus 2 right parenthesis end fraction end cell row blank equals cell 3 over 2 plus 5 over 6 plus 7 over 12 plus 9 over 20 plus 11 over 30 plus 13 over 42 end cell row blank equals cell fraction numerator 630 plus 350 plus 245 plus 189 plus 154 plus 130 over denominator 420 end fraction end cell row blank equals cell 1698 over 420 end cell row blank equals cell 283 over 70 end cell end table end style 

Jadi, nilai penjumlahan dari bentuk begin mathsize 14px style sum from p equals 0 to 5 of open parentheses fraction numerator 1 over denominator p plus 1 end fraction plus fraction numerator 1 over denominator p plus 2 end fraction close parentheses end style adalah begin mathsize 14px style 283 over 70 end style 

1

Roboguru

Diketahui:   Carilah  a.

Pembahasan Soal:

Ingat kembali mengenai definisi dan sifat notasi sigma:

Jika m less or equal than n semicolon space a subscript i space text bilangan real end text semicolon space m comma space n space text bilangan bulat end text semicolon dan c space text konstanta end text, maka:

  • sum from i equals m to n of a subscript i equals a subscript m plus a subscript m minus 1 end subscript plus a subscript m minus 2 end subscript plus midline horizontal ellipsis plus a subscript n 
  • sum from i equals m to n of c a subscript i equals c sum from i equals m to n of a subscript i  

Oleh karena itu, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 1 to 5 of 3 u subscript i end cell equals cell 3 sum from i equals 1 to 5 of u subscript i end cell row blank equals cell 3 open parentheses u subscript 1 plus u subscript 2 plus u subscript 3 plus u subscript 4 plus u subscript 5 close parentheses end cell row blank equals cell 3 open parentheses 4 plus 6 plus 8 plus 10 plus 12 close parentheses end cell row blank equals cell 3 open parentheses 40 close parentheses end cell row blank equals 120 end table   

 Dengan demikian, nilai dari sum from i equals 1 to 5 of 3 u subscript i adalah 120.space 

0

Roboguru

Nilai dari  adalah ...

Pembahasan Soal:

Dengan mensubstitusi nilai m, mulai dari m equals 1 sampai m equals 5 diperoleh 

Untuk space m equals 1 rightwards arrow fraction numerator 1 over denominator 1 plus 1 end fraction equals 1 half Untuk space m equals 2 rightwards arrow fraction numerator 2 over denominator 2 plus 1 end fraction equals 2 over 3 Untuk space m equals 3 rightwards arrow fraction numerator 3 over denominator 3 plus 1 end fraction equals 3 over 4 Untuk space m equals 4 rightwards arrow fraction numerator 4 over denominator 4 plus 1 end fraction equals 4 over 5 Untuk space m equals 5 rightwards arrow fraction numerator 5 over denominator 5 plus 1 end fraction equals 5 over 6 

Sehingga  

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from m equals 1 to 5 of fraction numerator m over denominator m plus 1 end fraction end cell equals cell 1 half plus 2 over 3 plus 3 over 4 plus 4 over 5 plus 5 over 6 end cell row blank equals cell fraction numerator 1 cross times 30 over denominator 2 cross times 30 end fraction plus fraction numerator 2 cross times 20 over denominator 3 cross times 20 end fraction plus fraction numerator 3 cross times 15 over denominator 4 cross times 15 end fraction plus fraction numerator 4 cross times 12 over denominator 5 cross times 12 end fraction plus fraction numerator 5 cross times 10 over denominator 6 cross times 10 end fraction end cell row blank equals cell 30 over 60 plus 40 over 60 plus 45 over 60 plus 48 over 60 plus 50 over 60 end cell row blank equals cell fraction numerator 30 plus 40 plus 45 plus 48 plus 50 over denominator 60 end fraction end cell row blank equals cell 213 over 60 end cell row blank equals cell fraction numerator 213 divided by 3 over denominator 60 divided by 3 end fraction end cell row blank equals cell 71 over 20 end cell row blank equals cell 60 over 20 plus 11 over 20 end cell row blank equals cell 3 plus 11 over 20 end cell row blank equals cell 3 11 over 20 end cell end table  

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Diketahui:   Carilah  e.

Pembahasan Soal:

Ingat kembali mengenai definisi dan sifat notasi sigma:

Jika m less or equal than n semicolon space a subscript i comma space b subscript i space text bilangan real; end text space m comma space n space text bilangan bulat end text, maka

  • sum from i equals m to n of a subscript i equals a subscript m plus a subscript m minus 1 end subscript plus a subscript m minus 2 end subscript plus midline horizontal ellipsis plus a subscript n
  • table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals m to n of open parentheses a subscript i plus b subscript i close parentheses open parentheses a subscript i minus b subscript i close parentheses end cell equals cell sum from i equals m to n of a subscript i squared minus sum from i equals m to n of b subscript i squared end cell end table 

Oleh karena itu, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 1 to 5 of open parentheses u subscript i plus v subscript i close parentheses left parenthesis u subscript i minus v subscript i right parenthesis end cell equals cell sum from i equals 1 to 5 of u subscript i squared minus sum from i equals 1 to 5 of v subscript i squared end cell row blank equals cell open parentheses u subscript 1 squared plus u subscript 2 squared plus u subscript 3 squared plus u subscript 4 squared plus u subscript 5 squared close parentheses end cell row blank blank cell negative open parentheses v subscript 1 squared plus v subscript 2 squared plus v subscript 3 squared plus v subscript 4 squared plus v subscript 5 squared close parentheses end cell row blank equals cell open parentheses 4 squared plus 6 squared plus 8 squared plus 10 squared plus 12 squared close parentheses end cell row blank blank cell negative open parentheses 1 squared plus 2 squared plus 4 squared plus 5 squared plus 6 squared close parentheses end cell row blank equals cell left parenthesis 16 plus 36 plus 64 plus 100 plus 144 right parenthesis end cell row blank blank cell negative left parenthesis 1 plus 4 plus 16 plus 25 plus 36 right parenthesis end cell row blank equals cell 360 minus 82 end cell row blank equals 278 end table   

 Dengan demikian, nilai dari sum from i equals 1 to 5 of open parentheses u subscript i plus v subscript i close parentheses open parentheses u subscript i minus v subscript i close parentheses adalah 278.space 

0

Roboguru

Pembahasan Soal:

Ingat bahwa notasi sigma didefinisikan sebagai berikut.

sum from i equals 1 to n of U subscript i equals U subscript 1 plus U subscript 2 plus U subscript 3 plus horizontal ellipsis plus U subscript n

Dengan:

  • i equals indeks blank penjumlahan
  • 1 equals batas blank bawah blank penjumlahan
  • n equals batas blank atas blank penjumlahan
  • open curly brackets 1 , 2 comma 3 comma horizontal ellipsis comma n close curly brackets equals wilayah blank penjumlahan

Dari definisi notasi sigma di atas maka akan didapatkan hasil sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from k equals 3 to 15 of open parentheses 4 k minus 7 close parentheses end cell equals cell open parentheses 4 cross times 3 minus 7 close parentheses plus open parentheses 4 cross times 4 minus 7 close parentheses plus open parentheses 4 cross times 5 minus 7 close parentheses plus open parentheses 4 cross times 6 minus 7 close parentheses end cell row blank blank cell plus open parentheses 4 cross times 7 minus 7 close parentheses plus open parentheses 4 cross times 8 minus 7 close parentheses plus open parentheses 4 cross times 9 minus 7 close parentheses plus open parentheses 4 cross times 10 minus 7 close parentheses end cell row blank blank cell plus open parentheses 4 cross times 11 minus 7 close parentheses plus open parentheses 4 cross times 12 minus 7 close parentheses plus open parentheses 4 cross times 13 minus 7 close parentheses end cell row blank blank cell plus open parentheses 4 cross times 14 minus 7 close parentheses plus open parentheses 4 cross times 15 minus 7 close parentheses end cell row blank equals cell open parentheses 12 minus 7 close parentheses plus open parentheses 16 minus 7 close parentheses plus open parentheses 20 minus 7 close parentheses plus open parentheses 24 minus 7 close parentheses plus open parentheses 28 minus 7 close parentheses end cell row blank blank cell plus open parentheses 32 minus 7 close parentheses plus open parentheses 36 minus 7 close parentheses plus open parentheses 40 minus 7 close parentheses plus open parentheses 44 minus 7 close parentheses end cell row blank blank cell plus open parentheses 48 minus 7 close parentheses plus open parentheses 52 minus 7 close parentheses plus open parentheses 56 minus 7 close parentheses plus open parentheses 60 minus 7 close parentheses end cell row blank equals cell 5 plus 9 plus 13 plus 17 plus 21 plus 25 plus 29 plus 33 plus 37 plus 41 end cell row blank blank cell plus 45 plus 59 plus 63 end cell row blank equals 397 end table end style

Jadi, dapat disimpulkan bahwa hasil dari sum from k equals 3 to 15 of open parentheses 4 k minus 7 close parentheses adalah 397.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved