Saving 9

An amount if $ 2000 is invested at an interest of 5% per month. if $ 200 is added at the beginning of each successive month but no withdrawals. Give an expression for the value accumulated after n months. After how many months will the amount have accumulated first exceed $ 42000.

Correct result:

n =  42

Solution:

$$\begin{gathered} x_0=2000 \\ {x}_1=x_0\cdot 1.05+200=2000\cdot 1.05+200=2300 \\ {x}_2=x_1\cdot 1.05+200=2300\cdot 1.05+200=2615 \\ {x}_3=x_2\cdot 1.05+200=2615\cdot 1.05+200={\displaystyle \frac{11783}{4}}=2945.75 \\ {x}_4=x_3\cdot 1.05+200=2945.75\cdot 1.05+200=3293.0375 \\ {x}_5=x_4\cdot 1.05+200=3293.0375\cdot 1.05+200\doteq 3657.6894 \\ {x}_6=x_5\cdot 1.05+200=3657.6894\cdot 1.05+200\doteq 4040.5738 \\ {x}_7=x_6\cdot 1.05+200=4040.5738\cdot 1.05+200\doteq 4442.6025 \\ {x}_8=x_7\cdot 1.05+200=4442.6025\cdot 1.05+200\doteq 4864.7327 \\ {x}_9=x_8\cdot 1.05+200=4864.7327\cdot 1.05+200\doteq 5307.9693 \\ {x}_{10}=x_9\cdot 1.05+200=5307.9693\cdot 1.05+200\doteq 5773.3678 \\ {x}_{11}=x_{10}\cdot 1.05+200=5773.3678\cdot 1.05+200\doteq 6262.0361 \\ {x}_{12}=x_{11}\cdot 1.05+200=6262.0361\cdot 1.05+200\doteq 6775.138 \\ {x}_{13}=x_{12}\cdot 1.05+200=6775.138\cdot 1.05+200\doteq 7313.8949 \\ {x}_{14}=x_{13}\cdot 1.05+200=7313.8949\cdot 1.05+200\doteq 7879.5896 \\ {x}_{15}=x_{14}\cdot 1.05+200=7879.5896\cdot 1.05+200\doteq 8473.5691 \\ {x}_{16}=x_{15}\cdot 1.05+200=8473.5691\cdot 1.05+200\doteq 9097.2475 \\ {x}_{17}=x_{16}\cdot 1.05+200=9097.2475\cdot 1.05+200\doteq 9752.1099 \\ {x}_{18}=x_{17}\cdot 1.05+200=9752.1099\cdot 1.05+200\doteq 10439.7154 \\ {x}_{19}=x_{18}\cdot 1.05+200=10439.7154\cdot 1.05+200\doteq 11161.7012 \\ {x}_{20}=x_{19}\cdot 1.05+200=11161.7012\cdot 1.05+200\doteq 11919.7862 \\ {x}_{21}=x_{20}\cdot 1.05+200=11919.7862\cdot 1.05+200\doteq 12715.7755 \\ {x}_{22}=x_{21}\cdot 1.05+200=12715.7755\cdot 1.05+200\doteq 13551.5643 \\ {x}_{23}=x_{22}\cdot 1.05+200=13551.5643\cdot 1.05+200\doteq 14429.1425 \\ {x}_{24}=x_{23}\cdot 1.05+200=14429.1425\cdot 1.05+200\doteq 15350.5997 \\ {x}_{25}=x_{24}\cdot 1.05+200=15350.5997\cdot 1.05+200\doteq 16318.1296 \\ {x}_{26}=x_{25}\cdot 1.05+200=16318.1296\cdot 1.05+200\doteq 17334.0361 \\ {x}_{27}=x_{26}\cdot 1.05+200=17334.0361\cdot 1.05+200\doteq 18400.7379 \\ {x}_{28}=x_{27}\cdot 1.05+200=18400.7379\cdot 1.05+200\doteq 19520.7748 \\ {x}_{29}=x_{28}\cdot 1.05+200=19520.7748\cdot 1.05+200\doteq 20696.8136 \\ {x}_{30}=x_{29}\cdot 1.05+200=20696.8136\cdot 1.05+200\doteq 21931.6543 \\ {x}_{31}=x_{30}\cdot 1.05+200=21931.6543\cdot 1.05+200\doteq 23228.237 \\ {x}_{32}=x_{31}\cdot 1.05+200=23228.237\cdot 1.05+200\doteq 24589.6488 \\ {x}_{33}=x_{32}\cdot 1.05+200=24589.6488\cdot 1.05+200\doteq 26019.1313 \\ {x}_{34}=x_{33}\cdot 1.05+200=26019.1313\cdot 1.05+200\doteq 27520.0878 \\ {x}_{35}=x_{34}\cdot 1.05+200=27520.0878\cdot 1.05+200\doteq 29096.0922 \\ {x}_{36}=x_{35}\cdot 1.05+200=29096.0922\cdot 1.05+200\doteq 30750.8968 \\ {x}_{37}=x_{36}\cdot 1.05+200=30750.8968\cdot 1.05+200\doteq 32488.4417 \\ {x}_{38}=x_{37}\cdot 1.05+200=32488.4417\cdot 1.05+200\doteq 34312.8637 \\ {x}_{39}=x_{38}\cdot 1.05+200=34312.8637\cdot 1.05+200\doteq 36228.5069 \\ {x}_{40}=x_{39}\cdot 1.05+200=36228.5069\cdot 1.05+200\doteq 38239.9323 \\ {x}_{41}=x_{40}\cdot 1.05+200=38239.9323\cdot 1.05+200\doteq 40351.9289 \\ {x}_{42}=x_{41}\cdot 1.05+200=40351.9289\cdot 1.05+200\doteq 42569.5253 \\ n=42 \end{gathered}$$

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