Asked by Anonymous
Starting from today, every month you save $500 into a bank account with earns interest at an annual rate of 4%. How much money do you have in the account at the end of 4 years?
My working is F=R(((1+i)^n)-1)-i where R=500, i=1/300 and n=48. Ans=$25,979.80
Am I solving it correctly? Thanks.
My working is F=R(((1+i)^n)-1)-i where R=500, i=1/300 and n=48. Ans=$25,979.80
Am I solving it correctly? Thanks.
Answers
Answered by
MathMate
You have assumed compound interest compounded monthly.
Assuming these assumptions are correct, then we have total amount F, R=monthly installment deposited at the <i>end</i> of the month, i=monthly interest, n=48, I=1+i,
F=R(I^(n-1)+I^(n-2)+...+I^(0))
=R(I^n-1)/(I-1)
=500((301/300)^48-1)/(1/300)
=25979.80
same as your answer, but your expression should have a division sign instead of the last negative sign.
If money is deposited at the <i>beginning</i> of the month, then
F=R(I^(n)+I^(n-1)+...+I^(1))
=RI(I^(n-1)+I^(n-2)+...+I^(0))
=RI(I^n-1)/(I-1)
=26066.40
Assuming these assumptions are correct, then we have total amount F, R=monthly installment deposited at the <i>end</i> of the month, i=monthly interest, n=48, I=1+i,
F=R(I^(n-1)+I^(n-2)+...+I^(0))
=R(I^n-1)/(I-1)
=500((301/300)^48-1)/(1/300)
=25979.80
same as your answer, but your expression should have a division sign instead of the last negative sign.
If money is deposited at the <i>beginning</i> of the month, then
F=R(I^(n)+I^(n-1)+...+I^(1))
=RI(I^(n-1)+I^(n-2)+...+I^(0))
=RI(I^n-1)/(I-1)
=26066.40
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