樣本試卷 DSE Maths MC Q39

At the beginning of the 1st month, amount = P
At the beginning of the 2nd month, amount = P(1 + 6%/12) + P
= 1.005P + P
At the beginning of the 3rd month, amount = (1.005P + P)(1 + 6%/12) + P
= 1.005^2P + 1.005P + P
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At the beginning of the 12th month, amount
= 1.005^11P + 1.005^10P + 1.005^9P + ...... + 1.005^2P + 1.005P + P
So at the end of the year, amount = amount at the beginning of the 12th month x (1 + 6^2/12)
= 1.005^12P + 1.005^11P + ....... + 1.005^2P + 1.005P
= P[1.005^12 + 1.005^11 + 1.005^10 + ...... + 1.005^2 + 1.005]
= P(1.005)[ 1.005^12 - 1]/(1.005 - 1)
= 12.3972P = 10,000
so P = 10,000/12.3972 = 806.63 (A)