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i=3∑8​(i+1)21​−i21​

Pertanyaan

sum from i equals 3 to 8 of space 1 over open parentheses i plus 1 close parentheses squared minus 1 over i squared 

Pembahasan Soal:

Gunakan konsep menentukan nilai dari notasi sigma.

Akan dtentukan nilai darisum from i equals 3 to 8 of space 1 over open parentheses i plus 1 close parentheses squared minus 1 over i squared.

Perhatikan perhitungan berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 2px end attributes row blank blank cell sum from i equals 3 to 8 of space 1 over open parentheses i plus 1 close parentheses squared minus 1 over i squared end cell row blank equals cell open parentheses 1 over open parentheses 3 plus 1 close parentheses squared minus 1 over open parentheses 3 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 4 plus 1 close parentheses squared minus 1 over open parentheses 4 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 5 plus 1 close parentheses squared minus 1 over open parentheses 5 close parentheses squared close parentheses plus end cell row blank blank cell open parentheses 1 over open parentheses 6 plus 1 close parentheses squared minus 1 over open parentheses 6 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 7 plus 1 close parentheses squared minus 1 over open parentheses 7 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 8 plus 1 close parentheses squared minus 1 over open parentheses 8 close parentheses squared close parentheses end cell row blank equals cell open parentheses 1 over open parentheses 4 close parentheses squared minus 1 over open parentheses 3 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 5 close parentheses squared minus 1 over open parentheses 4 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 6 close parentheses squared minus 1 over open parentheses 5 close parentheses squared close parentheses plus end cell row blank blank cell open parentheses 1 over open parentheses 7 close parentheses squared minus 1 over open parentheses 6 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 8 close parentheses squared minus 1 over open parentheses 7 close parentheses squared close parentheses plus open parentheses 1 over open parentheses 9 close parentheses squared minus 1 over open parentheses 8 close parentheses squared close parentheses end cell row blank equals cell 1 over open parentheses 4 close parentheses squared minus 1 over open parentheses 3 close parentheses squared plus 1 over open parentheses 5 close parentheses squared minus 1 over open parentheses 4 close parentheses squared plus 1 over open parentheses 6 close parentheses squared minus 1 over open parentheses 5 close parentheses squared end cell row blank blank cell 1 over open parentheses 7 close parentheses squared minus 1 over open parentheses 6 close parentheses squared plus 1 over open parentheses 8 close parentheses squared minus 1 over open parentheses 7 close parentheses squared plus 1 over open parentheses 9 close parentheses squared minus 1 over open parentheses 8 close parentheses squared end cell row blank equals cell negative 1 over open parentheses 3 close parentheses squared plus 1 over open parentheses 4 close parentheses squared minus 1 over open parentheses 4 close parentheses squared plus 1 over open parentheses 5 close parentheses squared minus 1 over open parentheses 5 close parentheses squared plus end cell row blank blank cell 1 over open parentheses 6 close parentheses squared minus 1 over open parentheses 6 close parentheses squared plus 1 over open parentheses 7 close parentheses squared minus 1 over open parentheses 7 close parentheses squared plus 1 over open parentheses 8 close parentheses squared minus 1 over open parentheses 8 close parentheses squared plus 1 over open parentheses 9 close parentheses squared end cell row blank equals cell negative 1 over open parentheses 3 close parentheses squared plus 1 over open parentheses 9 close parentheses squared end cell row blank equals cell negative 1 over 9 plus 1 over 81 end cell row blank equals cell fraction numerator negative 9 plus 1 over denominator 81 end fraction end cell row blank equals cell negative 8 over 81 end cell end table end style

Dengan demikian, diperoleh bahwa sum from i equals 3 to 8 of space 1 over open parentheses i plus 1 close parentheses squared minus 1 over i squared equals negative 8 over 81.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Dwi

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai dari  n=1∑3​3n−1n​ adalah ...

Pembahasan Soal:

Bentuk penjabaran darisum from n equals 1 to 3 of fraction numerator n over denominator 3 n minus 1 end fraction sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from n equals 1 to 3 of fraction numerator n over denominator 3 n minus 1 end fraction end cell equals cell fraction numerator 1 over denominator 3 minus 1 end fraction plus fraction numerator 2 over denominator 3 times 2 minus 1 end fraction plus fraction numerator 3 over denominator 3 times 3 minus 1 end fraction end cell row blank equals cell 1 half plus 2 over 5 plus 3 over 8 end cell row blank equals cell 0 comma 5 plus 0 comma 4 plus 0 comma 375 end cell row blank equals cell 1 comma 275 end cell end table

Jadi, jawaban yang tepat adalah E

 

0

Roboguru

Hasil dari n=1∑5​(n+n2) adalah ....

Pembahasan Soal:

Notasi sigma adalah notasi matematika untuk menuliskan suatu penjumlahan secara singkat.

i=1nf(i)=f(i=1)+f(i=2)+f(i=3)+...+f(i=n).

Berdasarkan definisi tersebut, maka:

n=15(n+n2)====(1+12)+(2+22)+(3+32)+(4+42)+(5+52)(1+1)+(2+4)+(3+9)+(4+16)+(5+25)2+6+12+20+3070

Jadi, hasil dari sum from n equals 1 to 5 of open parentheses n plus n squared close parentheses adalah 70.

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Diketahui i=−9∑−2​(k+i)=6. Bilangan bulat k yang memenuhi persamaan tersebut adalah...

Pembahasan Soal:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals negative 9 to negative 2 of open parentheses k plus i close parentheses end cell equals 6 row cell sum from i equals negative 9 to negative 2 of k plus sum from i equals negative 9 to negative 2 of i end cell equals 6 row cell k times 8 plus open square brackets negative 9 plus left parenthesis negative 8 right parenthesis plus left parenthesis negative 7 right parenthesis plus... plus left parenthesis negative 2 right parenthesis close square brackets end cell equals 6 row cell 8 k plus open square brackets 8 over 2 left parenthesis negative 9 plus left parenthesis negative 2 right parenthesis right parenthesis close square brackets end cell equals 6 row cell 8 k plus open square brackets 4 times left parenthesis negative 11 right parenthesis close square brackets end cell equals 6 row cell 8 k minus 44 end cell equals 6 row cell 8 k end cell equals cell 6 plus 44 end cell row cell 8 k end cell equals 50 row k equals cell 50 over 8 end cell row blank equals cell 25 over 4 space left parenthesis bukan space bilangan space bulat right parenthesis end cell end table end style

Jadi, tidak ada bilangan bulat  k yang memenuhi.

0

Roboguru

Tentukan hasil notasi sigma berikut : i=1∑4​(8i+5)+i=3∑6​(2i+3)

Pembahasan Soal:

Cari hasil dari notasi sigma tersebut dengan menjadikannya sebuah deret dengan mensubtitusikan nilai i equals 1 sampai i equals 4 ke fungsi 8 i plus 5 dan nilai  i equals 3 sampai i equals 6 ke fungsi 2 i plus 3 secara berurut seperti berikut

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from i equals 1 to 4 of space open parentheses 8 i plus 5 close parentheses plus sum from i equals 3 to 6 of space open parentheses 2 i plus 3 close parentheses end cell row blank equals cell open square brackets open parentheses 8 open parentheses 1 close parentheses plus 5 close parentheses plus open parentheses 8 open parentheses 2 close parentheses plus 5 close parentheses plus open parentheses 8 open parentheses 3 close parentheses plus 5 close parentheses plus left parenthesis 8 open parentheses 4 close parentheses plus 5 right parenthesis close square brackets end cell row blank blank cell plus open square brackets open parentheses 2 left parenthesis 3 right parenthesis plus 3 close parentheses plus open parentheses 2 left parenthesis 4 right parenthesis plus 3 close parentheses plus open parentheses 2 left parenthesis 5 right parenthesis plus 3 close parentheses plus open parentheses 2 left parenthesis 6 right parenthesis plus 3 close parentheses close square brackets end cell row blank equals cell open square brackets 13 plus 21 plus 29 plus 37 close square brackets plus open square brackets 9 plus 11 plus 13 plus 15 close square brackets end cell row blank equals cell 100 plus 48 end cell row blank equals 148 end table   

Jadi, hasil dari bentuk sum from i equals 1 to 4 of space open parentheses 8 i plus 5 close parentheses plus sum from i equals 3 to 6 of space open parentheses 2 i plus 3 close parentheses adalah 148.

0

Roboguru

Tentukan nilai x agar : p=x−1∑x+2​(2p+1)=60

Pembahasan Soal:

Substitusi nilai p equals x minus 1 comma space x comma space x plus 1 comma space x plus 2, sehingga

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from p equals x minus 1 to x plus 2 of left parenthesis 2 p plus 1 right parenthesis end cell equals 60 row cell open square brackets 2 left parenthesis x minus 1 right parenthesis plus 1 close square brackets plus open square brackets 2 left parenthesis x right parenthesis plus 1 close square brackets plus open square brackets 2 left parenthesis x plus 1 right parenthesis plus 1 close square brackets plus open square brackets 2 left parenthesis x plus 2 right parenthesis plus 1 close square brackets end cell equals 60 row cell 2 left parenthesis x up diagonal strike negative 1 end strike plus x plus x up diagonal strike plus 1 end strike plus x plus 2 right parenthesis plus 4 end cell equals 60 row cell 2 left parenthesis 4 x plus 2 right parenthesis plus 4 end cell equals 60 row cell 8 x plus 4 plus 4 end cell equals 60 row cell 8 x plus 8 end cell equals 60 row cell 8 x end cell equals 52 row x equals cell 52 over 8 end cell row blank equals cell 13 over 2 end cell end table end style

Jadi, diperoleh nilai table attributes columnalign right center left columnspacing 0px end attributes row x equals cell 13 over 2 end cell end table.

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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