Page 20 - E-Modul Olga Rosalinda Deret & Bilangan

5
b) (k 4)(k 2)
k 1

Penyelesaian:

5 5 5
a) (50 6k) = 50 6 k
k 1 k 1 k 1

= 5 . 50 – 6 (1 + 2 + 3 + 4 = 5)

= 250 - 90
= 160

5 5
2
b) (k 4)(k 2) = (k 2k 8)
k 1 k 1

5 5 5
2
= k 2 k 8
k 1 k 1 k 1
2
2
2
2
2
= (1 + 2 + 3 + 4 + 5 ) – 2(1 + 2 + 3 + 4 + 5) – 5 . 8
= 55 – 30 – 40
= -15

Contoh

Buktikanlah bahwa

n n n
2
2
(2k 7) 4 k 28 k 49 n
k 1 k 1 k 1
Bukti:

n n
2
2
(2k 7) (4k 28k 49)
k 1 k 1

n n n
2
4k 28k 49
k 1 k 1 k 1

n n
2
4 k 28 k 49 n
k 1 k 1

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