Question #173528

3b) A bank pays you 4.5% interest per year. In addition, you receive |100 as bonus at

the end of the year (after the interest is paid). Find a recurrence for the amount of

money after n years if you invest |2000.


Expert's answer

Let r=0.045 is an interest rate, S0=2000 the initial amount of money, Sn - the amount of money after n years, and b=100 is the annual bonus. Then we have a recurrent equation:

Sn=(1+r)Sn−1+bS_n=(1+r)S_{n-1}+b

Sn−Sn−1=r(Sn−1+b/r)S_n-S_{n-1}=r(S_{n-1}+b/r)

(Sn+b/r)−(Sn−1+b/r)=r(Sn−1+b/r)(S_{n}+b/r)-(S_{n-1}+b/r)=r(S_{n-1}+b/r)

Sn+b/r=(1+r)(Sn−1+b/r)S_{n}+b/r=(1+r)(S_{n-1}+b/r)

The last equation shows that the sequence Sn+b/rS_{n}+b/r is a geometric progression, and we can write

Sn+b/r=(1+r)n(S0+b/r)S_{n}+b/r=(1+r)^n(S_{0}+b/r)

Sn=(1+r)n(S0+b/r)−b/rS_{n}=(1+r)^n(S_{0}+b/r)-b/r

Sn=(1+0.045)n(2000+100/0.045)−100/0.045=4222.22⋅1.045n−2222.22S_{n}=(1+0.045)^n(2000+100/0.045)-100/0.045=4222.22\cdot1.045^n-2222.22

Answer. Sn=4222.22⋅1.045n−2222.22S_{n}=4222.22\cdot1.045^n-2222.22


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