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Hasil dari begin mathsize 14px style sum from n equals 4 to 25 of open parentheses n plus 7 close parentheses equals... end style 

Pembahasan Video:

Pembahasan Soal:

Diketahui begin mathsize 14px style sum from n equals 4 to 25 of left parenthesis n plus 7 right parenthesis end style. Dengan memanfaatkan sifat notasi sigma

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

akibatnya, didapat persamaan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from n equals 4 to 25 of left parenthesis n plus 7 right parenthesis end cell equals cell sum from n equals 4 minus 3 to 25 minus 3 of open parentheses open parentheses n plus 3 close parentheses plus 7 close parentheses space end cell row blank equals cell sum from n equals 1 to 22 of left parenthesis n plus 10 right parenthesis space end cell end table end style 

Dengan demikian, hasil dari  begin mathsize 14px style sum from n equals 4 to 25 of left parenthesis n plus 7 right parenthesis end style sama dengan begin mathsize 14px style sum from n equals 1 to 22 of left parenthesis n plus 10 right parenthesis end style

Jadi, jawaban yang tepat adalah C.

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Terakhir diupdate 03 Mei 2021

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Pembahasan Soal:

Dengan menggunakan sifat notasi sigma 

sum from i equals m to n of space u subscript i equals sum from i equals m minus p to n minus p of space u subscript i plus p end subscript

sehingga 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sum from k equals 7 to 10 of left parenthesis k minus 6 right parenthesis end cell equals cell sum from k equals 7 minus 6 to 10 minus 6 of left parenthesis k plus 6 minus 6 right parenthesis end cell row blank equals cell sum from k equals 1 to 4 of k end cell end table end style  

 

Jadi, bentuk table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from k equals 7 to 10 of left parenthesis k minus 6 right parenthesis end cell end table setara dengan table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from k equals 1 to 4 of k end cell end table .

0

Roboguru

Jika dinyatakan dengan batas bawah 1, bentuk menjadi...

Pembahasan Soal:

Ingat sIfat notasi sigma

begin mathsize 14px style sum from k equals m to n of a subscript k equals sum from k equals m minus p to n minus p of a subscript k plus p end subscript equals sum from k equals m plus p to n plus p of a subscript k minus p end subscript end style .

 

maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from k equals 0 to 5 of left parenthesis 2 k plus 1 right parenthesis plus sum from k equals 7 to 10 of left parenthesis 2 k minus 1 right parenthesis end cell equals cell sum from k equals 0 plus 1 to 5 plus 1 of left parenthesis 2 left parenthesis k minus 1 right parenthesis plus 1 right parenthesis plus sum from k equals 7 minus 6 to 10 minus 6 of left parenthesis 2 left parenthesis k plus 6 right parenthesis minus 1 right parenthesis end cell row blank equals cell sum from k equals 1 to 6 of left parenthesis 2 k minus 2 plus 1 right parenthesis plus sum from k equals 1 to 4 of left parenthesis 2 k plus 12 minus 1 right parenthesis end cell row blank equals cell sum from k equals 1 to 6 of left parenthesis 2 k minus 1 right parenthesis plus sum from k equals 1 to 4 of left parenthesis 2 k plus 11 right parenthesis end cell end table 

 

Jadi, bentuk table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from k equals 0 to 5 of left parenthesis 2 k plus 1 right parenthesis plus sum from k equals 7 to 10 of left parenthesis 2 k minus 1 right parenthesis end cell end table jika dinyatakan dengan batas bawah 1 adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from k equals 1 to 6 of left parenthesis 2 k minus 1 right parenthesis plus sum from k equals 1 to 4 of left parenthesis 2 k plus 11 right parenthesis end cell end table.

 

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.

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Buktikan bahwa:

Pembahasan Soal:

begin mathsize 14px style sum from k plus 1 to n of k squared plus sum from k equals 4 to n plus 3 of left parenthesis 2 k plus 1 right parenthesis equals sum from k plus 1 to n of k squared plus sum from k equals 4 minus 3 to n plus 3 minus 3 of left parenthesis 2 left parenthesis k plus 3 right parenthesis plus 1 right parenthesis equals sum from k plus 1 to n of k squared plus sum from k equals 1 to n of left parenthesis 2 k plus 6 plus 1 right parenthesis equals sum from k plus 1 to n of k squared plus sum from k equals 1 to n of left parenthesis 2 k plus 7 right parenthesis equals sum from k plus 1 to n of left parenthesis k squared plus 2 k plus 7 right parenthesis subscript left parenthesis Terbukti right parenthesis end subscript space end style 

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Roboguru

Nilai n yang memenuhi  adalah...

Pembahasan Soal:

begin mathsize 14px style sum from k equals 1 to n of left parenthesis 4 k plus 5 right parenthesis equals 774 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 4 cross times 1 plus 5 right parenthesis plus left parenthesis 4 cross times 2 plus 5 right parenthesis plus left parenthesis 4 cross times 3 plus 5 right parenthesis plus... plus left parenthesis 4 cross times n plus 5 right parenthesis end cell equals 774 row cell 9 plus 13 plus 17 plus 21 plus... plus left parenthesis 4 n plus 5 right parenthesis end cell equals 774 end table end style 

Notasi sigma di atas merupakan deret aritmatika dengan begin mathsize 14px style a equals 9 comma space U subscript n equals left parenthesis 4 n plus 5 right parenthesis comma space dan space S subscript n equals 774 end style

Maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell S subscript n end cell equals cell n over 2 open parentheses a plus U subscript n close parentheses end cell row 774 equals cell n over 2 left parenthesis 9 plus left parenthesis 4 n plus 5 right parenthesis right parenthesis end cell row 774 equals cell n over 2 left parenthesis 14 plus 4 n right parenthesis end cell row 774 equals cell n left parenthesis 7 plus 2 n right parenthesis end cell row 774 equals cell 7 n plus 2 n squared end cell row cell 2 n squared plus 7 n minus 774 end cell equals 0 row cell left parenthesis n minus 18 right parenthesis left parenthesis n plus 43 right parenthesis end cell equals 0 row blank blank blank end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis n minus 18 right parenthesis end cell equals cell 0 space space space a t a u space left parenthesis 2 n plus 43 right parenthesis equals 0 end cell row n equals cell 18 space a t a u space space space space space space space space space space space space space space n equals negative 43 over 2 subscript left parenthesis tidak space memenuhi right parenthesis end subscript end cell end table end style 

Jadi, jawaban yang tepat adalah B. 

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