Question #200987

Find the accumulated amount after 4 years if $1000 is invested at 6% per year compounded (a) daily (assume a 365-day year) and (b) continuously...Find the accumulated amount after 4 years if $1000 is invested at 6% per year compounded (a) annually, (b) semiannually, (c) quarterly, (d) monthly, and (e) daily.


1
Expert's answer
2021-06-02T13:57:00-0400

1.

(a) Compound interest


A=P(1+rn)ntA=P(1+\dfrac{r}{n})^{nt}

Daily: n=365n=365


A=1000(1+0.06365)365(4)=$1271.22A=1000(1+\dfrac{0.06}{365})^{365(4)}=\$1271.22

(b) Continuously Compounded Interest


A=Pert=1000e0.06(4)=$1271.25A=Pe^{rt}=1000e^{0.06(4)}=\$1271.25

2.

Compound interest


A=P(1+rn)ntA=P(1+\dfrac{r}{n})^{nt}

(a) Annually n=1n=1


A=1000(1+0.061)1(4)=$1262.48A=1000(1+\dfrac{0.06}{1})^{1(4)}=\$1262.48

(b) Semiannually n=2n=2


A=1000(1+0.062)2(4)=$1266.77A=1000(1+\dfrac{0.06}{2})^{2(4)}=\$1266.77


(c) Quarterly n=4n=4


A=1000(1+0.064)4(4)=$1268.99A=1000(1+\dfrac{0.06}{4})^{4(4)}=\$1268.99



(d) Monthly n=12n=12


A=1000(1+0.0612)12(4)=$1270.49A=1000(1+\dfrac{0.06}{12})^{12(4)}=\$1270.49



(e) Daily: n=365n=365


A=1000(1+0.06365)365(4)=$1271.22A=1000(1+\dfrac{0.06}{365})^{365(4)}=\$1271.22




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