Asked by fizz
At the end of every year for 3 years, RM1000 will be invested in an account that offers 8% compounded annualy.Find the account amount at the end of the 3 years.
the answer is $ 3246.40
in my calculation i solve
1000(1+o.08/1)^(3-1)
=11664
the answer is $ 3246.40
in my calculation i solve
1000(1+o.08/1)^(3-1)
=11664
Answers
Answered by
Steve
what you have done is calculate the first year's growth.
Now you have to add the 2nd year:
1000(1+.08)^(2-1) = 1080.00
Then the third year: 1000
1166.40+1080.00+1000.00 = 3246.40
As you can see, for n years, you would have
1000(1+1.08+1.08^2 + ... + 1.08^(n-1))
= 1000(1.08^n - 1)/(1.08-1)
which for n=3, is
1000(1.08^3 - 1)/(1.08-1) = 3246.40
Now you have to add the 2nd year:
1000(1+.08)^(2-1) = 1080.00
Then the third year: 1000
1166.40+1080.00+1000.00 = 3246.40
As you can see, for n years, you would have
1000(1+1.08+1.08^2 + ... + 1.08^(n-1))
= 1000(1.08^n - 1)/(1.08-1)
which for n=3, is
1000(1.08^3 - 1)/(1.08-1) = 3246.40
Answered by
Reiny
Just use the formula.
Amount = payment( (1+i)^n -1)/i
Amount = 1000(1.08)^3 -1)/.08
= 3246.40
Amount = payment( (1+i)^n -1)/i
Amount = 1000(1.08)^3 -1)/.08
= 3246.40
Answered by
fizz
ok thanks both of you
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