Сalculate according to the formula
"A=S\\times(i+\\frac{i}{(1+i)^n-1})"
"1000=S\\times(0.001+\\frac{0.001}{(1+0.001)^{30}-1})"
S=29 540
"2000=S\\times(0.001+\\frac{0.001}{(1+0.001)^{30}-1})"
S=59 080
"3000=S\\times(0.001+\\frac{0.001}{(1+0.001)^{30}-1})"
S=88 620
"4000=S\\times(0.001+\\frac{0.001}{(1+0.001)^{30}-1})"
S=118 160
"5000=S\\times(0.001+\\frac{0.001}{(1+0.001)^{30}-1})"
S=147 700
Yes, the payment amount is doubled and the total amount is doubled. similarly, and if the payment amount increases by 3 times, then the total amount by times increases and so on, and so on.
It turns out a direct proportional relationship between the payment amount and the total amount
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