Roboguru

Tentukan deret 12+32+52+⋯ sampai dengan n suku ....

Pertanyaan

Tentukan deret 1 squared plus 3 squared plus 5 squared plus midline horizontal ellipsis sampai dengan n suku ....

  1. fraction numerator n left parenthesis 3 n plus 2 right parenthesis left parenthesis 2 n minus 1 right parenthesis over denominator 3 end fraction 

  2. fraction numerator n left parenthesis 2 n plus 2 right parenthesis left parenthesis 2 n minus 1 right parenthesis over denominator 3 end fraction

  3. fraction numerator 2 n left parenthesis n plus 1 right parenthesis left parenthesis n minus 1 right parenthesis over denominator 3 end fraction

  4. fraction numerator 2 left parenthesis n plus 1 right parenthesis left parenthesis 2 n minus 1 right parenthesis over denominator 2 end fraction

  5. fraction numerator 2 plus left parenthesis n minus 1 right parenthesis left parenthesis 16 plus 8 left parenthesis n minus 2 right parenthesis right parenthesis over denominator 2 end fraction

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.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Kurnia

Mahasiswa/Alumni Universitas Jember

Terakhir diupdate 07 Oktober 2021

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Pertanyaan yang serupa

Tentukan deret 12+32+52+⋯ 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

Diketahui ∑k=26​(nk−5)2=335, jika nilai n positif, nilai 2n2−3n adalah ....

Pembahasan Soal:

Diketahui : begin inline style sum from k equals 2 to 6 of end style left parenthesis n k minus 5 right parenthesis squared equals 335 dan nilai n positif

DItanya : nilai 2 straight n squared minus 3 straight n ?

Jawab :
Ingat sifat notasi sigma:

begin inline style sum from k equals p to q of end style space A subscript left parenthesis k right parenthesis end subscript equals begin inline style sum from k equals p minus a to q minus a of end style space A subscript left parenthesis k plus a right parenthesis end subscript

Dengan sifat tersebut, maka bentuk sum from k equals 2 to 6 of left parenthesis n k minus 5 right parenthesis squared equals 335 dapat diubah menjadi:

begin inline style sum from k equals 2 minus 1 to 6 minus 1 of end style open parentheses n left parenthesis k plus 1 right parenthesis minus 5 close parentheses squared equals 335 begin inline style sum from k equals 1 to 5 of end style begin inline style open parentheses n left parenthesis k plus 1 right parenthesis minus 5 close parentheses squared end style equals 335 begin inline style sum from k equals 1 to 5 of end style begin inline style open parentheses n k plus n minus 5 close parentheses squared end style begin inline style equals end style begin inline style 335 end style begin inline style sum from k equals 1 to 5 of end style begin inline style space end style begin inline style left parenthesis end style begin inline style n end style begin inline style k end style begin inline style plus end style begin inline style n end style begin inline style minus end style begin inline style 5 end style begin inline style right parenthesis end style begin inline style left parenthesis end style begin inline style n end style begin inline style k end style begin inline style plus end style begin inline style n end style begin inline style minus end style begin inline style 5 end style begin inline style right parenthesis end style begin inline style equals end style begin inline style 335 end style begin inline style sum from k equals 1 to 5 of space left parenthesis n squared k squared plus n squared k minus 5 n k plus n squared k plus n squared minus 5 n minus 5 n k minus 5 n plus 25 right parenthesis equals 335 end style begin inline style sum from k equals 1 to 5 of end style begin inline style space end style begin inline style left parenthesis end style begin inline style n squared end style begin inline style k squared end style begin inline style plus end style begin inline style 2 end style begin inline style n squared end style begin inline style k end style begin inline style minus end style begin inline style 10 end style begin inline style n end style begin inline style k end style begin inline style minus end style begin inline style 10 end style begin inline style n end style begin inline style plus end style begin inline style n squared end style begin inline style plus end style begin inline style 25 end style begin inline style right parenthesis end style begin inline style equals end style begin inline style 335 end style

Ingat sifat pemisahan notasi sigma:

begin inline style sum from k equals p to q of end style open parentheses A subscript left parenthesis k right parenthesis end subscript plus B subscript left parenthesis k right parenthesis end subscript close parentheses equals begin inline style sum from k equals p to q of end style space A subscript left parenthesis k right parenthesis end subscript plus begin inline style sum from k equals p to q of end style space B subscript left parenthesis k right parenthesis end subscript

Dengan sifat tersebut, maka bentuk sum from k equals 1 to 5 of begin inline style space end style begin inline style left parenthesis end style begin inline style n squared end style begin inline style k squared end style begin inline style plus end style begin inline style 2 end style begin inline style n squared end style begin inline style k end style begin inline style minus end style begin inline style 10 end style begin inline style n end style begin inline style k end style begin inline style minus end style begin inline style 10 end style begin inline style n end style begin inline style plus end style begin inline style n squared end style begin inline style plus end style begin inline style 25 end style begin inline style right parenthesis end style begin inline style equals end style begin inline style 335 end style dapat dipisah menjadi:

begin inline style sum from k equals 1 to 5 of end style space n squared k squared plus begin inline style sum from k equals 1 to 5 of end style space 2 n squared k minus begin inline style sum from k equals 1 to 5 of end style space 10 n k minus begin inline style sum from k equals 1 to 5 of end style space 10 n plus begin inline style sum from k equals 1 to 5 of space n squared end style plus begin inline style sum from k equals 1 to 5 of end style space 25 equals 335

Ingat sifat notasi sigma:

begin inline style sum from k equals p to q of end style space r times A subscript left parenthesis k right parenthesis end subscript equals r times begin inline style sum from k equals p to q of end style space A subscript left parenthesis k right parenthesis end subscript, dengan r konstanta.

Karena n konstanta, maka:

space n squared times begin inline style sum from k equals 1 to 5 of end style space k squared plus 2 n squared times begin inline style sum from k equals 1 to 5 of end style space k minus 10 n times begin inline style sum from k equals 1 to 5 of end style space k minus begin inline style sum from k equals 1 to 5 of end style space 10 n plus begin inline style sum from k equals 1 to 5 of space n squared end style plus begin inline style sum from k equals 1 to 5 of end style space 25 equals 335

Ingat sifat notasi sigma:

1. begin inline style sum from k equals 1 to p of end style space k equals 1 half p left parenthesis p plus 1 right parenthesis

2. begin inline style sum from k equals 1 to p of end style space k squared equals 1 over 6 p left parenthesis p plus 1 right parenthesis left parenthesis 2 p plus 1 right parenthesis

3. begin inline style sum from k equals 1 to p of end style space n equals n times p comma space n space konstanta

Dari ketiga sifat tersebut, maka didapat:

table attributes columnalign right center left columnspacing 0px end attributes row cell n squared times 1 over 6 left parenthesis 5 right parenthesis left parenthesis 5 plus 1 right parenthesis left parenthesis 2 times 5 plus 1 right parenthesis plus 2 n squared times 1 half left parenthesis 5 right parenthesis left parenthesis 5 plus 1 right parenthesis minus 10 n times 1 half left parenthesis 5 right parenthesis left parenthesis 5 plus 1 right parenthesis minus 10 n times 5 plus n squared times 5 plus 25 times 5 end cell equals 335 row cell 55 n squared plus 30 n squared minus 150 n minus 50 n plus 5 n squared plus 125 end cell equals 335 row cell 90 n squared minus 200 n minus 210 end cell equals 0 row cell 9 n squared minus 20 n minus 21 end cell equals 0 end table

Nilai n dari persamaan kuadrat tersebut dengan koefisien-koefisien a equals 9 comma space b equals negative 20 comma dan c equals negative 21 dapat ditentukan dengan rumus:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight n subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative straight b plus-or-minus square root of straight b squared minus 4 straight a cross times straight c end root over denominator 2 straight a end fraction end cell row blank equals cell fraction numerator 20 plus-or-minus square root of left parenthesis negative 20 right parenthesis squared minus 4 open parentheses 9 close parentheses cross times open parentheses negative 21 close parentheses end root over denominator 2 cross times 9 end fraction end cell row blank equals cell fraction numerator 20 plus-or-minus square root of 400 plus 756 end root over denominator 18 end fraction end cell row blank equals cell fraction numerator 20 plus-or-minus square root of 1.156 end root over denominator 18 end fraction end cell row blank equals cell fraction numerator 20 plus-or-minus 34 over denominator 18 end fraction end cell end table

Karena nilai n positif, maka n yang diambil adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight n subscript 1 end cell equals cell fraction numerator 20 plus 34 over denominator 18 end fraction end cell row blank equals cell 54 over 18 end cell row blank equals 3 end table

Lalu substitusi nilai n ke 2 straight n squared minus 3 straight n.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight n squared minus 3 straight n end cell equals cell 2 open parentheses 3 close parentheses squared minus 3 open parentheses 3 close parentheses end cell row blank equals cell 18 minus 9 end cell row blank equals 9 end table

Dengan demikian, nilai 2 straight n squared minus 3 straight n adalah 9.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Nilai t=1∑4​(3t2+4t)=…

Pembahasan Soal:

Ingat pada sifat pada notasi sigma, jika terdapat sum from i equals 1 to n of a subscript i plus sum from i equals 1 to n of b subscript i maka menjadi  sum from i equals 1 to n of left parenthesis a subscript i plus b subscript i right parenthesis, sehingga

sum from t equals 1 to 4 of open parentheses 3 t squared plus 4 t close parentheses equals sum from t equals 1 to 4 of 3 t squared plus sum from t equals 1 to 4 of 4 t equals 3 sum from t equals 1 to 4 of t squared plus 4 sum from t equals 1 to 4 of t

Perhatikan bahwa,

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from t equals 1 to 4 of t squared end cell equals cell 1 squared plus 2 squared plus 3 squared plus 4 squared end cell row blank equals cell 1 plus 4 plus 9 plus 16 end cell row blank equals 30 end table

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from t equals 1 to 4 of t end cell equals cell 1 plus 2 plus 3 plus 4 end cell row blank equals 10 end table

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell stack 3 sum with t equals 1 below and 4 on top t squared plus 4 sum from t equals 1 to 4 of t end cell equals cell left parenthesis 3 cross times 30 right parenthesis plus left parenthesis 4 cross times 10 right parenthesis end cell row blank equals cell 90 plus 40 end cell row blank equals 130 end table

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Nilai dari n=0∑3​(n2−2n+1)=....

Pembahasan Soal:

Dengan memperhatikan aturan pemisahan sigma dalam beberapa penjumlahan, maka didapatkan

sum from n equals 0 to 3 of open parentheses n squared minus 2 n plus 1 close parentheses equals sum from n equals 0 to 3 of n squared plus sum from n equals 0 to 3 of left parenthesis negative 2 n right parenthesis plus sum from n equals 0 to 3 of 1 equals sum from n equals 0 to 3 of n squared minus sum from n equals 0 to 3 of 2 n plus sum from n equals 0 to 3 of 1

Dengan melakukan substitusi nilai n yang bergerak dari 0 sampai 3, didapatkan

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from n equals 0 to 3 of n squared end cell equals cell 0 squared plus 1 squared plus 2 squared plus 3 squared end cell row blank equals cell 0 plus 1 plus 4 plus 9 end cell row blank equals 14 row cell sum from n equals 0 to 3 of 2 n end cell equals cell 2 open parentheses 0 close parentheses plus 2 open parentheses 1 close parentheses plus 2 open parentheses 2 close parentheses plus 2 open parentheses 3 close parentheses end cell row blank equals cell 0 plus 2 plus 4 plus 6 end cell row blank equals 12 row cell sum from n equals 0 to 3 of 1 end cell equals cell 4 cross times 1 end cell row blank equals 4 end table

Jadi, sum from n equals 0 to 3 of open parentheses n squared minus 2 n plus 1 close parentheses equals 6

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Nilai dari n=3∑9​(n2−n)=....

Pembahasan Soal:

Ingat bahwa

sum from n equals a to b of open parentheses x subscript n plus y subscript n close parentheses equals sum from n equals a to b of x subscript n plus sum from n equals a to b of y subscript n

maka berdasarkan persamaan diatas dan definisi notasi sigma, didapatkan

sum from n equals 3 to 9 of open parentheses n squared minus n close parentheses equals sum from n equals 3 to 9 of n squared minus sum from n equals 3 to 9 of n equals open parentheses 3 squared plus 4 squared plus 5 squared plus 6 squared plus 7 squared plus 8 squared plus 9 squared close parentheses space space space space minus open parentheses 3 plus 4 plus 5 plus 6 plus 7 plus 8 plus 9 close parentheses equals open parentheses 9 plus 16 plus 25 plus 36 plus 49 plus 64 plus 81 close parentheses minus 42 equals 280 minus 42 equals 238

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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