1.2.3+2.3.4+3.4.5+...+8.9.10=

1 . 2 . 3 + 2 . 3 . 4 + 3 . 4 . 5 + 4 . 5 . 6 +  ... + 8 . 9 . 10 = ?
n = 1
n + 1 = 2
n + 2 = 3

maka, polanya = n . (n + 1) . (n + 2)

n . (n + 1) . (n + 2)
= n . (n² + 3n + 2)
= n³ + 3n² + 2n

Karena tiga bilangan berurutan dari 1 sampai 8, maka 

1
∑
n = 8

**********************

1
∑ (n³ + 3n² + 2n)
n = 8

    1              1              1
 = ∑ n³ + 3 . ∑ n² + 2 . ∑ n
   n = 8        n = 8       n = 8

= (1³ + 2³ + 3³ + ... + 8³) + 3 . (1² + 2² + 3² + ... + 8²) + 2 . (1 + 2 + 3 .. + 8)

*** (1³ + 2³ + .. + n³) = (n/2 . (n + 1))²
*** (1³ + 2³ + .. + 8³) = (8/2 . 9)²
*** (1³ + 2³ + .. + 8³) = 36²
*** (1³ + 2³ + .. + 8³) = 1.296

*** (1² + 2² + 3² + ... + n²) = 1/6 . n (n + 1) . (2n + 1)
*** (1² + 2² + 3² + ... + 8²) = 1/6 . 8 . (8 + 1) . (2 . 8 + 1)
*** (1² + 2² + 3² + ... + 8²) = 1/6 . 8 . 9 . 17
*** (1² + 2² + 3² + ... + 8²) = 204
maka, 
3 . (1² + 2² + 3² + ... + 8²) 
= 3 . 204
= 612

*** 1 + 2 + 3 + ... + n = n/2 . (n + 1)
*** (1 + 2 + 3 .. + 8) = 8/2 . 9
*** (1 + 2 + 3 .. + 8) = 36
maka,
2 . (1 + 2 + 3 .. + 8)
= 2 . 36
= 72

(1³ + 2³ + 3³ + ... + 8³) + 3 . (1² + 2² + 3² + ... + 8²) + 2 . (1 + 2 + 3 .. + 8)
= 1.296 + 612 + 72
= 1.980

Kata kunci : ∑, n, n², n³
Kode soal : 12.2.7
Kelas 12
Pelajaran Matematika
Bab 7 Barisan dan Deret 

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