论文部分内容阅读
第(52)题的答案: 分析,开始第—个月需垫付资金260元。第二个月因已收回25元,故只垫付250-25=225元。第三个月又收回25元,只垫付225-25=200元。如此类推,直到第十个月只垫付25元。十个月垫付资金之和为250+225+200+…………+25是个递减等差数列。首项a_1=250,末项a_(?)=25,项数n=10,S_(?)是n项等差数列的代数和,亦即垫付资金总额。S为贴息额。∴垫付资金总额:S_(?)=(a_1+a_(?))n/2=(250+25)×10/2=275×10/2=2,750/2=1,375元。因垫付资金而贴息额:S=1,375×6‰=1,375×6/1000=8.25元。
The answer to (52): Analysis, the first month to be paid 260 yuan. The second month because has been recovered 25 yuan, it is only advance 250-25 = 225 yuan. The third month and recovered 25 yuan, only advance 225-25 = 200 yuan. And so on, until the tenth month only advance 25 yuan. Ten months to pay the sum of funds for the 250 +225 +200 + ............ +25 is a descending differential series. The first a_1 = 250, the last a _ (?) = 25, the number of items n = 10, S_ (?) Is the algebraic sum of the arithmetic progression of n terms, that is, advances the total amount of funds. S is the discount amount. Total funds advanced: S _ (?) = (A_1 + a _ (?)) N / 2 = (250 + 25) × 10/2 = 275 × 10/2 = 2,750/2 = 1,375 yuan. Due to advance payment of funds and interest rates: S = 1,375 × 6 ‰ = 1,375 × 6/1000 = 8.25 yuan.