Jawaban:
Nilai dari \left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= 7\frac{1}{20}
20
1
. Simak pembahasan mengenai notasi sigma berikut.
Pembahasan
Rumus umum notasi sigma adalah
\left \sum {{n} \atop {i=1}} \right. f(x_{i})f(x
i
) = f(x_{1})f(x
1
) + f(x_{2})f(x
2
) + f(x_{3})f(x
3
) + .... + f(x_{n})f(x
n
)
Maka nilai dari \left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
adalah
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= \frac{(-2)^{2}}{(-2) + 3}
(−2)+3
(−2)
2
+ \frac{(-1)^{2}}{(-1) + 3}
(−1)+3
(−1)
2
+ \frac{0^{2}}{0 + 3}
0+3
0
2
+ \frac{1^{2}}{1 + 3}
1+3
1
2
+ \frac{2^{2}}{2 + 3}
2+3
2
2
+ \frac{3^{2}}{3 + 3}
3+3
3
2
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= \frac{4}{1}
1
4
+ \frac{1}{2}
2
1
+ \frac{0}{3}
3
0
+ \frac{1}{4}
4
1
+ \frac{4}{5}
5
4
+ \frac{9}{6}
6
9
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= 4 + \frac{1}{2}
2
1
+ 0 + \frac{1}{4}
4
1
+ \frac{4}{5}
5
4
+ \frac{3}{2}
2
3
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= 4 + \frac{1}{2}
2
1
+ \frac{1}{4}
4
1
+ \frac{4}{5}
5
4
+ \frac{3}{2}
2
3
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= \frac{80}{20}
20
80
+ \frac{10}{20}
20
10
+ \frac{5}{20}
20
5
+ \frac{16}{20}
20
16
+ \frac{30}{20}
20
30
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= \frac{80 + 10 + 5 + 16 + 30}{20}
20
80+10+5+16+30
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= \frac{141}{20}
20
141
\left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
= 7\frac{1}{20}
20
1
Jadi nilai dari \left \sum {{3} \atop {k=-2}} \right. \frac{k^{2}}{k + 3}
k+3
k
2
adalah 7\frac{1}{20}
20
1
Pelajari lebih lanjut
Menyatakan deret geometri dalam bentuk notasi sigma https://brainly.co.id/tugas/23213682
Menentukan banyak pola pada suku ke- n deret geometri https://brainly.co.id/tugas/23191664
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Detil jawaban
Kelas: 11
Mapel: Matematika
Bab: Notasi sigma
Kode: 11.2.
Kata kunci: notasi sigma, jumlah, nilai
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