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Ubahlah notasi sigma berikut menjadi notasi sigma dengan batas atas 12. b.

Pertanyaan

Ubahlah notasi sigma berikut menjadi notasi sigma dengan batas atas 12.

b. sum from i equals 12 to 18 of open parentheses i squared plus 1 close parentheses

Pembahasan Soal:

Perhatikan sifat notasi sigma berikut.

sum from k equals c to n of U subscript k equals sum from k equals c minus m to n minus m of U subscript k plus m end subscript 

Sehingga notasi sigma sum from i equals 12 to 18 of open parentheses i squared plus 1 close parentheses dapat dinyatakan dengan batas atas 12 sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 12 to 18 of open parentheses i squared plus 1 close parentheses end cell equals cell sum from i equals 12 minus 6 to 18 minus 6 of open parentheses open parentheses i plus 6 close parentheses squared plus 1 close parentheses end cell row blank equals cell sum from i equals 6 to 12 of open parentheses i squared plus 12 i plus 36 plus 1 close parentheses end cell row blank equals cell sum from i equals 6 to 12 of open parentheses i squared plus 12 i plus 37 close parentheses end cell end table 

Dengan demikian, diperoleh table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 12 to 18 of open parentheses i squared plus 1 close parentheses end cell equals cell sum from i equals 6 to 12 of open parentheses i squared plus 12 i plus 37 close parentheses end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Afrisno

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 12 Agustus 2021

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Pertanyaan yang serupa

Diketahui:   Maka nilai  yang memenuhi adalah ....

Pembahasan Soal:

Dengan menerapkan sifat notasi sigma sum from straight k equals straight n to straight m of straight k equals sum from straight k equals straight n plus straight p to straight m plus straight p of left parenthesis straight k minus straight p right parenthesis maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 4 minus straight x to 8 minus straight x of left parenthesis 4 straight k plus 1 right parenthesis end cell equals cell stack sum 4 left parenthesis straight k minus straight x right parenthesis plus 1 with straight k equals 4 minus straight x plus straight x below and 8 minus straight x plus straight x on top end cell row blank equals cell sum from straight k equals 4 to 8 of 4 left parenthesis straight k minus straight x right parenthesis plus 1 end cell row blank equals cell 4 left parenthesis 4 minus straight x right parenthesis plus 1 plus 4 left parenthesis 5 minus straight x right parenthesis plus 1 plus 4 left parenthesis 6 minus straight x right parenthesis plus 1 end cell row blank blank cell plus 4 left parenthesis 7 minus straight x right parenthesis plus 1 plus 4 left parenthesis 8 minus straight x right parenthesis plus 1 end cell row blank equals cell 16 minus 4 straight x plus 1 plus 20 minus 4 straight x plus 1 plus 24 minus 4 straight x plus 1 end cell row blank blank cell plus 28 minus 4 straight x plus 1 plus 32 minus 4 straight x plus 1 end cell row blank equals cell 125 minus 20 straight x end cell end table 

Diketahui bahwa table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from straight k equals 4 minus straight x to 8 minus straight x of left parenthesis 4 straight k plus 1 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 85 end table maka: 

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from straight k equals 4 minus straight x to 8 minus straight x of left parenthesis 4 straight k plus 1 right parenthesis end cell equals 85 row cell 125 minus 20 straight x end cell equals 85 row cell negative 20 straight x end cell equals cell negative 40 end cell row straight x equals 2 end table 

Jadi, Nilai straight x yang table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from straight k equals 4 minus straight x to 8 minus straight x of left parenthesis 4 straight k plus 1 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 85 end table memenuhi adalah x equals 2. 

Oleh karena itu, Jawaban yang benar adalah A

 

0

Roboguru

Notasi sigma yang ekuivalen dengan  adalah ...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell sum from k equals 5 to 12 of open parentheses 2 minus 5 k close parentheses squared end cell equals cell sum from k equals 5 plus 3 to 12 plus 3 of open parentheses 2 minus 5 open parentheses k minus 3 close parentheses close parentheses squared end cell row blank equals cell sum from k equals 8 to 15 of open parentheses 2 minus 5 k plus 15 close parentheses squared end cell row blank equals cell sum from k equals 8 to 15 of open parentheses 17 minus 5 k close parentheses squared end cell end table end style 

Jadi, jawaban yang tepat adalah D

0

Roboguru

Bentuk setara dengan bentuk ....

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 14px style sum from q equals 6 to 20 of q squared equals 6 squared plus 7 squared plus 8 squared plus 9 squared plus 10 squared plus horizontal ellipsis plus 20 squared equals open square brackets 6 squared plus 7 squared plus 8 squared plus 9 squared close square brackets plus open square brackets 10 squared plus horizontal ellipsis plus 20 squared close square brackets equals sum from q equals 6 to 9 of q squared plus sum from q equals 10 to 20 of q squared end style

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Hasil dari  adalah ….

Pembahasan Soal:

Ingat kembali sifat pada notasi sigma berikut ini!

begin mathsize 14px style sum from k equals 1 to n of U subscript k equals sum from k equals 1 minus p to n minus p of open parentheses U subscript k plus p close parentheses end style 

Kemudian, perhatikan juga sifat notasi sigma berikut ini!

begin mathsize 14px style sum from k equals 1 to m of U subscript k plus sum from k equals 1 to m of V subscript k equals sum from k equals 1 to m of open parentheses U subscript k plus V subscript k close parentheses end style 

Sehingga, didapat perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from k equals 1 to 8 of open parentheses 4 k minus 3 close parentheses plus sum from k equals 1 to 4 of open parentheses 5 k plus 2 close parentheses end cell row blank equals cell sum from k equals 1 to 4 of open parentheses 4 k minus 3 close parentheses plus sum from k equals 5 to 8 of open parentheses 4 k minus 3 close parentheses plus sum from k equals 1 to 4 of open parentheses 5 k plus 2 close parentheses end cell row blank equals cell open parentheses sum from k equals 1 to 4 of open parentheses 4 k minus 3 close parentheses plus sum from k equals 1 to 4 of open parentheses 5 k plus 2 close parentheses close parentheses plus sum from k equals 5 to 8 of open parentheses 4 k minus 3 close parentheses end cell row blank equals cell sum from k equals 1 to 4 of open parentheses open parentheses 4 k minus 3 close parentheses plus open parentheses 5 k plus 2 close parentheses close parentheses plus sum from k equals 5 to 8 of open parentheses 4 k minus 3 close parentheses end cell row blank equals cell sum from k equals 1 to 4 of open parentheses 9 k minus 1 close parentheses plus sum from k equals 5 to 8 of open parentheses 4 k minus 3 close parentheses end cell row blank equals cell sum from k equals 1 to 4 of open parentheses 9 k minus 1 close parentheses plus sum from k equals 5 minus 4 to 8 minus 4 of open parentheses 4 open parentheses k plus 4 close parentheses minus 3 close parentheses end cell row blank equals cell sum from k equals 1 to 4 of open parentheses 9 k minus 1 close parentheses plus sum from k equals 1 to 4 of open parentheses 4 k plus 13 close parentheses end cell row blank equals cell sum from k equals 1 to 4 of open parentheses open parentheses 9 k minus 1 close parentheses plus open parentheses 4 k plus 13 close parentheses close parentheses end cell row blank equals cell sum from k equals 1 to 4 of open parentheses 13 k plus 12 close parentheses end cell row blank blank blank end table end style 

Dengan demikian, hasil dari begin mathsize 14px style sum from k equals 1 to 8 of open parentheses 4 k minus 3 close parentheses plus sum from k equals 1 to 4 of open parentheses 5 k plus 2 close parentheses end style adalah begin mathsize 14px style sum from k equals 1 to 4 of open parentheses 13 k plus 12 close parentheses end style.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Ubahlah menjadi batas bawah 4:

Pembahasan Soal:

Pada soal di atas kita asumsikan bahwa n equals k.

Ingat rumus sum from p equals 1 to q of u subscript p equals sum from p equals 1 plus r to q plus r of u subscript p minus r end subscript

Berdasarkan di atas maka diperoleh:

sum from k equals 1 to 20 of open parentheses k squared plus 2 k plus 1 close parentheses equals sum from k equals 1 plus 3 to 20 plus 3 of open parentheses open parentheses k minus 3 close parentheses squared plus 2 open parentheses k minus 3 close parentheses plus 1 close parentheses

                                    equals sum from k equals 4 to 23 of open parentheses open parentheses k squared minus 6 k plus 9 close parentheses plus open parentheses 2 k minus 6 close parentheses plus 1 close parentheses

                               equals sum from k equals 4 to 23 of open parentheses k squared minus 6 k plus 9 plus 2 k minus 6 plus 1 close parentheses

         equals sum from k equals 4 to 23 of open parentheses k squared minus 4 k plus 4 close parentheses

Dengan demikian,  sum from n equals 1 to 20 of open parentheses k squared plus 2 k plus 1 close parentheses equals sum from k equals 4 to 23 of open parentheses k squared minus 4 k plus 4 close parentheses

0

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