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標題:
Geometric Sequence
發問:
Plan A for five yearsWith the first deposit (at the end of first month) of $1000,$1000 is deposited each month.0.04% of the total deposit (per month) guaranteed profit is added ti the sum as a bonus for continuation.The total deposit will ve returned after fivr years.Find the total earned at the end of... 顯示更多 Plan A for five years With the first deposit (at the end of first month) of $1000,$1000 is deposited each month.0.04% of the total deposit (per month) guaranteed profit is added ti the sum as a bonus for continuation.The total deposit will ve returned after fivr years. Find the total earned at the end of first year whan adopting plan A.
At the end of the first month, total = $1000 At the end of the second month, total = 1000(1 + 0.04) + 1000 = 1000(1.04) + 1000 At the end of the third month, total = 1000(1 + 0.04)2 + 1000(1 + 0.04) + 1000 = 1000(1.04)2 + 1000(1.04) + 1000 Hence, at the end of the first year, total = 1000(1.04)11 + 1000(1.04)10 + ... + 1000 = 1000[1 + (1.04) + ... + (1.04)11] = 1000[(1)(1.0412 - 1) / (1.04 - 1)] = $15 025.81 (cor. to the nearest cent)
其他解答:
Geometric Sequence
發問:
Plan A for five yearsWith the first deposit (at the end of first month) of $1000,$1000 is deposited each month.0.04% of the total deposit (per month) guaranteed profit is added ti the sum as a bonus for continuation.The total deposit will ve returned after fivr years.Find the total earned at the end of... 顯示更多 Plan A for five years With the first deposit (at the end of first month) of $1000,$1000 is deposited each month.0.04% of the total deposit (per month) guaranteed profit is added ti the sum as a bonus for continuation.The total deposit will ve returned after fivr years. Find the total earned at the end of first year whan adopting plan A.
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最佳解答:At the end of the first month, total = $1000 At the end of the second month, total = 1000(1 + 0.04) + 1000 = 1000(1.04) + 1000 At the end of the third month, total = 1000(1 + 0.04)2 + 1000(1 + 0.04) + 1000 = 1000(1.04)2 + 1000(1.04) + 1000 Hence, at the end of the first year, total = 1000(1.04)11 + 1000(1.04)10 + ... + 1000 = 1000[1 + (1.04) + ... + (1.04)11] = 1000[(1)(1.0412 - 1) / (1.04 - 1)] = $15 025.81 (cor. to the nearest cent)
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