Roboguru

Notasi sigma yang menyatakan  adalah...

Pertanyaan

Notasi sigma yang menyatakan begin mathsize 14px style negative 1 plus 2 plus 9 plus 20 plus... plus 1.377 end style adalah...

  1. begin mathsize 14px style sum from k equals 2 to 27 of left parenthesis 2 k squared minus 3 k minus 3 right parenthesis end style

  2. size 14px sum from size 14px k size 14px equals size 14px 3 to size 14px 27 of size 14px left parenthesis size 14px k to the power of size 14px 2 size 14px minus size 14px 3 size 14px k size 14px minus size 14px 1 size 14px right parenthesis 

  3. size 14px sum from size 14px k size 14px equals size 14px 1 to size 14px 27 of size 14px left parenthesis size 14px 2 size 14px k begin mathsize 14px style left parenthesis k minus 3 right parenthesis end style size 14px right parenthesis 

  4. size 14px sum from size 14px k size 14px equals size 14px 1 to size 14px 27 of size 14px left parenthesis size 14px k begin mathsize 14px style left parenthesis 2 k minus 3 k right parenthesis end style size 14px right parenthesis 

  5. size 14px sum from size 14px k size 14px equals size 14px 1 to size 14px 27 of size 14px left parenthesis size 14px 2 size 14px k to the power of size 14px 2 size 14px minus size 14px 3 size 14px k size 14px right parenthesis 

Pembahasan Video:

Pembahasan Soal:

Deret bilangan di atas merupakan deret bilangan bertingkat dua.

begin mathsize 14px style U subscript n equals a n squared plus b n plus c end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell U subscript 1 end cell equals cell negative 1 end cell row cell a cross times 1 squared plus b cross times 1 plus c end cell equals cell negative 1 end cell row cell a plus b plus c end cell equals cell negative 1 space space space... space left parenthesis 1 right parenthesis end cell row cell U subscript 2 end cell equals 2 row cell a cross times 2 squared plus b cross times 2 plus c end cell equals 2 row cell 4 a plus 2 b plus c end cell equals cell 2 space space space space space space space... space left parenthesis 2 right parenthesis end cell row cell U subscript 3 end cell equals 9 row cell a cross times 3 squared plus b cross times 3 plus c end cell equals 9 row cell 9 a plus 3 b plus c end cell equals cell 9 space space space space space space space... space left parenthesis 3 right parenthesis end cell end table end style 

Eliminasi begin mathsize 14px style c end style dari persamaan begin mathsize 14px style left parenthesis 1 right parenthesis space dan space left parenthesis 2 right parenthesis end style

begin mathsize 14px style bottom enclose table attributes columnalign right center left columnspacing 2px end attributes row cell a plus b plus c end cell equals cell negative 1 end cell row cell 4 a plus 2 b plus c end cell equals 2 end table end enclose subscript long dash space space space space space space minus 3 a minus b equals negative 3 space space space space space space space space space space space space... space left parenthesis 4 right parenthesis end style    

Eliminasi begin mathsize 14px style c end style dari persamaan begin mathsize 14px style left parenthesis 2 right parenthesis space dan space left parenthesis 3 right parenthesis end style

begin mathsize 14px style bottom enclose table attributes columnalign left center left columnspacing 2px 2px 2px end attributes row cell 4 a plus 2 b plus c end cell equals 2 row cell 9 a plus 3 b plus c end cell equals 9 end table end enclose subscript long dash space space space space space space space minus 5 a minus b equals minus 7 space space space space space space space space space space space space... space left parenthesis 5 right parenthesis end style 

Eliminasi begin mathsize 14px style b end style dari persamaan begin mathsize 14px style left parenthesis 4 right parenthesis space dan space left parenthesis 5 right parenthesis end style:

begin mathsize 14px style bottom enclose table attributes columnalign right center left columnspacing 2px end attributes row cell negative 3 a minus b end cell equals cell negative 3 end cell row cell negative 5 a minus b end cell equals cell negative 7 end cell end table end enclose subscript long dash space end subscript space space space space space space space space space space space space 2 a equals 4 space space space space space space space space space space space space space a equals 2 end style 

Substitusi begin mathsize 14px style a equals 2 end style ke dalam persamaan undefined

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 a minus b end cell equals cell negative 3 end cell row cell negative 3 cross times 2 minus b end cell equals cell negative 3 end cell row cell negative 6 minus b end cell equals cell negative 3 end cell row cell negative b end cell equals 3 row b equals cell negative 3 end cell end table end style 

Substitusikan undefined  ke dalam persamaan begin mathsize 14px style left parenthesis 1 right parenthesis end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a plus b plus c end cell equals cell negative 1 end cell row cell 2 plus left parenthesis negative 3 right parenthesis plus c end cell equals cell negative 1 end cell row cell negative 1 plus c end cell equals cell negative 1 end cell row c equals 0 end table end style 

Diperoleh begin mathsize 14px style a equals 2 comma space b equals negative 3 comma space dan space c equals 0 end style

Rumus suku ke-n begin mathsize 14px style open parentheses U subscript n close parentheses end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell U subscript n end cell equals cell a n squared plus b n plus c end cell row cell U subscript n end cell equals cell 2 cross times n squared plus left parenthesis negative 3 right parenthesis cross times n plus 0 end cell row cell U subscript n end cell equals cell 2 n squared minus 3 n end cell end table end style 

Batas bawah sigma adalah undefined, karena begin mathsize 14px style negative 1 end style merupakan suku pertama deret.

Menentukan batas atas sigma:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell U subscript n end cell equals cell 2 n squared minus 3 n end cell row cell 1.377 end cell equals cell 2 n squared minus 3 n end cell row cell 2 n squared minus 3 n minus 1.377 end cell equals 0 row cell left parenthesis n minus 27 right parenthesis left parenthesis 2 n plus 51 right parenthesis end cell equals 0 row cell left parenthesis n minus 27 right parenthesis space atau space left parenthesis 2 straight n plus 51 right parenthesis end cell equals 0 end table end style 

begin mathsize 14px style n equals 27 space atau space straight n equals negative 51 over 2 subscript left parenthesis tidak space memenuhi right parenthesis end subscript end style 

Diperoleh batas atas begin mathsize 14px style n equals 27 end style.

Sehingga, notasi sigma dari deret tersebut adalah begin mathsize 14px style sum from k equals 1 to 27 of open parentheses 2 k squared minus 3 k close parentheses end style

Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Freelancer6

Terakhir diupdate 03 Mei 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 deret  sampai dengan n suku ....

Pembahasan Soal:

Dengan menggunakan aturan dan sifat notasi sigma maka didapatkan

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 squared plus 3 squared plus 5 squared plus midline horizontal ellipsis end cell row blank equals cell sum from i equals 1 to n of left parenthesis 2 i minus 1 right parenthesis squared end cell row blank equals cell sum from i equals 1 to n of 4 i squared minus 4 i plus 1 end cell row blank equals cell 4 sum from i equals 1 to n of i squared minus 4 sum from i equals 1 to n of i plus sum from i equals 1 to n of 1 end cell row blank equals cell 4 open square brackets fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction close square brackets minus 4 open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets plus n end cell row blank equals cell fraction numerator 2 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 3 end fraction minus 2 n left parenthesis n plus 1 right parenthesis plus n end cell row blank equals cell n over 3 open square brackets 2 open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses minus 6 open parentheses n plus 1 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 2 open parentheses n plus 1 close parentheses open parentheses 2 n minus 2 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 open parentheses n plus 1 close parentheses open parentheses n minus 1 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 open parentheses n squared minus 1 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 n squared minus 4 plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 n squared minus 1 close square brackets end cell row blank equals cell fraction numerator n left parenthesis 2 n plus 1 right parenthesis left parenthesis 2 n minus 1 right parenthesis over denominator 3 end fraction end cell end table

Oleh karena itu, tidak ada jawaban yang benar pada Opsi.

0

Roboguru

Tentukan deret  sampai dengan n suku ....

Pembahasan Soal:

Dengan menggunakan aturan dan sifat notasi sigma maka didapatkan

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 squared plus 3 squared plus 5 squared plus midline horizontal ellipsis end cell equals cell sum from i equals 1 to n of left parenthesis 2 i minus 1 right parenthesis squared end cell row blank equals cell sum from i equals 1 to n of 4 i squared minus 4 i plus 1 end cell row blank equals cell 4 sum from i equals 1 to n of i squared minus 4 sum from i equals 1 to n of i plus sum from i equals 1 to n of 1 end cell row blank equals cell 4 open square brackets fraction numerator n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 6 end fraction close square brackets minus 4 open square brackets fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets plus n end cell row blank equals cell fraction numerator 2 n open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses over denominator 3 end fraction minus 2 n left parenthesis n plus 1 right parenthesis plus n end cell row blank equals cell n over 3 open square brackets 2 open parentheses n plus 1 close parentheses open parentheses 2 n plus 1 close parentheses minus 6 open parentheses n plus 1 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 2 open parentheses n plus 1 close parentheses open parentheses 2 n minus 2 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 open parentheses n plus 1 close parentheses open parentheses n minus 1 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 open parentheses n squared minus 1 close parentheses plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 n squared minus 4 plus 3 close square brackets end cell row blank equals cell n over 3 open square brackets 4 n squared minus 1 close square brackets end cell row blank equals cell fraction numerator n left parenthesis 2 n plus 1 right parenthesis left parenthesis 2 n minus 1 right parenthesis over denominator 3 end fraction end cell end table

Oleh karena itu, tidak ada jawaban yang benar pada Opsi.

0

Roboguru

Tentukan nilai dari

Pembahasan Soal:

Dengan memperhatikan definisi notasi sigma, maka

1 squared plus 2 squared plus 3 squared plus midline horizontal ellipsis plus 20 squared equals sum from n equals 1 to 20 of n squared

Ingat bahwa 

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from n equals 1 to 20 of n 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 blank rightwards arrow cell sum from n equals 1 to 20 of n squared equals fraction numerator 20 open parentheses 21 close parentheses open parentheses 41 close parentheses over denominator 6 end fraction end cell row blank equals cell 10 cross times 7 cross times 41 end cell row blank equals cell 2.870 end cell end table

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

Nilai dari begin mathsize 14px style sum from k equals 3 to 7 of open parentheses k squared plus open parentheses k minus 3 close parentheses squared close parentheses end style adalah

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from k equals 3 to 7 of open parentheses k squared plus open parentheses k minus 3 close parentheses squared close parentheses end cell equals cell open parentheses 3 squared plus open parentheses 3 minus 3 close parentheses squared close parentheses plus open parentheses 4 squared plus open parentheses 4 minus 3 close parentheses squared close parentheses plus open parentheses 5 squared plus open parentheses 5 minus 3 close parentheses squared close parentheses plus open parentheses 6 squared plus open parentheses 6 minus 3 close parentheses squared close parentheses plus open parentheses 7 squared plus open parentheses 7 minus 3 close parentheses squared close parentheses end cell row blank equals cell open parentheses 3 squared plus 0 squared close parentheses plus open parentheses 4 squared plus 1 squared close parentheses plus open parentheses 5 squared plus 2 squared close parentheses plus open parentheses 6 squared plus 3 squared close parentheses plus open parentheses 7 squared plus 4 squared close parentheses end cell row blank equals cell open parentheses 9 plus 0 close parentheses plus open parentheses 16 plus 1 close parentheses plus open parentheses 25 plus 4 close parentheses plus open parentheses 36 plus 9 close parentheses plus open parentheses 49 plus 16 close parentheses end cell row blank equals cell 9 plus 17 plus 29 plus 45 plus 65 end cell row blank equals 165 end table end style 

Jadi, tidak ada pilihan jawaban yang tepat.

 

0

Roboguru

Buktikan bahwa:

Pembahasan Soal:

begin mathsize 14px style sum from k equals 10 to 20 of left parenthesis k squared plus 20 k plus 103 right parenthesis plus sum from k equals 10 to n minus 11 of left parenthesis k squared plus 42 k plus 444 right parenthesis equals sum from k equals 10 plus 10 to 20 plus 10 of left parenthesis left parenthesis k minus 10 right parenthesis squared plus 20 left parenthesis k minus 10 right parenthesis plus 103 right parenthesis space space space space space space plus sum from k equals 10 plus 21 to n minus 11 plus 21 of left parenthesis left parenthesis k minus 21 right parenthesis squared plus 42 left parenthesis k minus 21 right parenthesis plus 444 right parenthesis equals sum from k equals 20 to 30 of left parenthesis k squared minus 20 k plus 100 plus 20 k minus 200 plus 103 right parenthesis space space space space space space plus sum from k equals 31 to n plus 10 of left parenthesis k squared minus 42 k plus 441 plus 42 k minus 882 plus 444 right parenthesis equals sum from k equals 20 to 30 of left parenthesis k squared plus 3 right parenthesis plus sum from k equals 31 to n plus 10 of left parenthesis k squared plus 3 right parenthesis equals sum from k equals 20 to n plus 10 of left parenthesis k squared plus 3 right parenthesis subscript left parenthesis Terbukti right parenthesis end subscript end style   

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