Pertanyaan

i = 1 ∑ n ( 2 i + 5 ) = n ( n + 6 )

$i = 1 \sum n (2 i + 5) = n (n + 6)$

G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Pembahasan

Ingatlah! Maka, a. Gunakan aturan perkalian konstan, Maka, b. Gunakan aturan, Maka, Di dapatkan, Jadi, terbukti seperti pada penjelasan di atas.

Ingatlah!

sum a subscript n plus b subscript n equals sum a subscript n plus sum b subscript n

Maka,

sum from i equals 1 to n of 2 i plus 5 equals sum from i equals 1 to n of 2 i plus sum from i equals 1 to n of 5

a. sum from i equals 1 to n of 2 i

Gunakan aturan perkalian konstan,

sum for blank of c times a subscript n equals c times sum for blank of a subscript n

Maka,

$table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 1 to n of 2 i end cell equals cell 2 times sum from i equals 1 to n of i end cell row cell begin inline style sum from k equals 1 to n of end style k end cell equals cell 1 half n left parenthesis n plus 1 right parenthesis end cell row cell 2 times sum from i equals 1 to n of i end cell equals cell 1 half n open parentheses n plus 1 close parentheses end cell row blank equals cell fraction numerator 1 times thin space 2 over denominator 2 end fraction n open parentheses n plus 1 close parentheses end cell row blank equals cell open parentheses n plus 1 close parentheses n times thin space 1 end cell row blank equals cell n open parentheses n plus 1 close parentheses end cell end table$

b. sum from i equals 1 to n of 5

Gunakan aturan,

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from k equals 1 to n of end cell equals cell 1 space to the power of a equals a times n end exponent end cell row a equals 5 end table

Maka,

a times n equals 5 n

Di dapatkan,

Jadi, sum from i equals 1 to n of open parentheses 2 i plus 5 close parentheses equals n left parenthesis n plus 6 right parenthesis terbukti seperti pada penjelasan di atas.