Jawab:
[tex]\sum \limits_{k\:=\:2}^{5} (k + 1)^2 = 86[/tex]
Koreksi:
Mungkin ada kesalahan dalam soal, yang dimana tertulis di soal
[tex]\sum \limits_{k\:=\:2}^{5} (k = 1)^2[/tex] , mungkin seharusnya tertulis [tex]\sum \limits_{k\:=\:2}^{5} (k + 1)^2[/tex]
Rumus:
[tex]\sum \limits_{i\:=\:y}^{n} f(x) = f(y) + f(y +1)+f(y+2)+...+f(n)[/tex]
[tex]\sum \limits_{i\:=\:x}^{n} a \:f(x) = a \sum \limits_{i\:=\:x}^{n} f(x)[/tex]
[tex]\sum \limits_{i\:=\:x}^{n} f(x) + g(x) = \sum \limits_{i\:=\:x}^{n} f(x) + \sum \limits_{i\:=\:x}^{n} g(x)[/tex]
[tex]\sum \limits_{i\:=\:1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}[/tex]
[tex]\sum \limits_{i\:=\:x}^{n} (a+b)^2 = \sum \limits_{i\:=\:x}^{n} (a^2 + 2ab + b^2)[/tex]
[tex]\sum \limits_{i\:=\:x}^{n} (a+b)^2 = \sum \limits_{i\:=\:x}^{n} a^2 + \sum \limits_{i\:=\:x}^{n} 2ab + \sum \limits_{i\:=\:x}^{n} b^2[/tex]
[tex]\sum \limits_{i\:=\:x}^{n} (a+b)^2 = \sum \limits_{i\:=\:x}^{n} a^2 + 2\sum \limits_{i\:=\:x}^{n} ab + \sum \limits_{i\:=\:x}^{n} b^2[/tex]
[tex]\sum \limits_{i\:=\:1}^{n} k = kn[/tex]
[tex]\sum \limits_{i\:=\:x}^{n} k = \sum \limits_{i\:=\:(x-1)}^{n-1} k[/tex]
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