Jika Sn merupakan jumlah n suku pertama suatu deret aritmetika, buktikan bahwa: Sn+3 – 3Sn+2 + 3Sn+1 – Sn = 0
Jika Sn merupakan jumlah n suku pertama suatu deret aritmetika, buktikan bahwa: Sn+3 – 3Sn+2 + 3Sn+1 – Sn = 0
Pembahasan:
Sn+3 – 3Sn+2 + 3Sn+1 – Sn = 0
Sn+3 – Sn – 3Sn+2 + 3Sn+1 = 0
Sn+3 – Sn – 3(Sn+2 – Sn+1) = 0
(Un+3 + Un+2 + Un+1 + Sn) – Sn – 3(Un+2 + Sn+1 – Sn+1) = 0
Un+3 + Un+2 + Un+1 – 3Un+2= 0
Un+3 + Un+1 – 2Un+2 = 0
(Un+3 – Un+2) + (Un+1 – Un+2) = 0
b – b = 0
(terbukti)
Pembahasan:
Sn+3 – 3Sn+2 + 3Sn+1 – Sn = 0
Sn+3 – Sn – 3Sn+2 + 3Sn+1 = 0
Sn+3 – Sn – 3(Sn+2 – Sn+1) = 0
(Un+3 + Un+2 + Un+1 + Sn) – Sn – 3(Un+2 + Sn+1 – Sn+1) = 0
Un+3 + Un+2 + Un+1 – 3Un+2= 0
Un+3 + Un+1 – 2Un+2 = 0
(Un+3 – Un+2) + (Un+1 – Un+2) = 0
b – b = 0
(terbukti)
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