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Nilai dari ∑ i = 4 51 ​ ( 4 i − 5 ) = ... .

Nilai dari .

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diperoleh nilai dari .

diperoleh nilai dari begin inline style sum from i equals 4 to 51 of end style space open parentheses 4 i minus 5 close parentheses equals 5.040.

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Ditanyanilai dari . Ingat sifat notasi sigma: Maka, Ingat sifat notasi sigma: Dengan sifat tersebut maka Ingat sifat notasi sigma: Dengan sifat tersebut didapat: Ingat sifat notasi sigma: dan Dengan sifat tersebut, didapat: Jadi, diperoleh nilai dari .

Ditanya nilai dari begin inline style sum from i equals 4 to 51 of end style space open parentheses 4 i minus 5 close parentheses.

Ingat sifat notasi sigma:

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

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 4 to 51 of space open parentheses 4 i minus 5 close parentheses end cell equals cell begin inline style sum from i equals 4 minus 3 to 51 minus 3 of end style space open parentheses 4 left parenthesis i plus 3 right parenthesis minus 5 close parentheses end cell row blank equals cell begin inline style sum from i equals 1 to 48 of end style space open parentheses 4 i plus 12 minus 5 close parentheses end cell row blank equals cell begin inline style sum from i equals 1 to 48 of end style space left parenthesis 4 i plus 7 right parenthesis end cell end table

Ingat sifat notasi sigma:

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

Dengan sifat tersebut maka

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from i equals 1 to 48 of space left parenthesis 4 i plus 7 right parenthesis end cell end table equals begin inline style sum from i equals 1 to 48 of end style space 4 i plus begin inline style sum from i equals 1 to 48 of end style space 7

Ingat sifat notasi sigma:

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

Dengan sifat tersebut didapat:

sum from i equals 1 to 48 of space 4 i plus begin inline style sum from i equals 1 to 48 of end style space 7 equals 4 times begin inline style sum from i equals 1 to 48 of end style space i plus begin inline style sum from i equals 1 to 48 of end style space 7

Ingat sifat notasi sigma:

begin inline style sum from i equals 1 to n of end style space i equals 1 half n left parenthesis n plus 1 right parenthesis

dan

begin inline style sum from i equals 1 to n of end style space k equals n times k

Dengan sifat tersebut, didapat:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 times begin inline style sum from i equals 1 to 48 of end style space i plus begin inline style sum from i equals 1 to 48 of end style space 7 end cell equals cell 4 times 1 half left parenthesis 48 right parenthesis left parenthesis 48 plus 1 right parenthesis plus left parenthesis 48 right parenthesis left parenthesis 7 right parenthesis end cell row blank equals cell 2 left parenthesis 48 right parenthesis left parenthesis 49 right parenthesis plus left parenthesis 48 right parenthesis left parenthesis 7 right parenthesis end cell row blank equals cell 4.704 plus 336 end cell row blank equals cell 5.040 end cell end table

Jadi, diperoleh nilai dari begin inline style sum from i equals 4 to 51 of end style space open parentheses 4 i minus 5 close parentheses equals 5.040.

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