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Mathematics 25 Online
OpenStudy (anonymous):

Suppose on December 31, 2000, a deposit of $100 is made in a savings account that pays 10% annual interest. So one year after the initial deposit, on December 31, 2001, the account will be credited with $10, and have a value of $110. On December 31, 2002 that account will be credited with an additional $11, and have value $121. Find a recursive relation that gives the value of the account n years after the initial deposit.

OpenStudy (amistre64):

i always find it helpful to write out a few "terms" of the sequence to see if a pattern develops

OpenStudy (amistre64):

b1 = (100*.1) + 11 b2 = b1*.1 + 11 = ((100*.1 +11) *.1) + 11 etc

OpenStudy (amistre64):

1.1 is a better fit

OpenStudy (amistre64):

b3 = b2*1.1 + 11 = ((100*1.1 +11) *1.1 + 11)*1.1 + 11 = (100*1.1^2 + 11*1.1 + 11)* 1.1 + 11 = 100*1.1^3 + 11*1.1^2 + 11*1.1 + 11 = 100*1.1^3 + 11 (1.1^2 + 1.1 + 1) so to follow a pattern emerging I would say that: b4 = 100(1.1)^4 + 11(1.1^3 + 1.1^2 + 1.1 + 1) b5 = 100(1.1)^5 + 11(1.1^4 + 1.1^3 + 1.1^2 + 1.1 + 1) etc

OpenStudy (amistre64):

since we are looking for the nth term, we need to make this more general \[b_r = 100(1.1)^r + 11(1.1^{r-1} + 1.1^{r-2} + 1.1^{r-3}+...+1.1^3+1.1^2 + 1.1 + 1)\] that last part is a geometric series

OpenStudy (anonymous):

It looks right, does this satisfy the definition of recursiveness?

OpenStudy (anonymous):

Being that the last part is a geometric series, would it be better to write it in sigma summation notation?

OpenStudy (amistre64):

im thinking this is turning into explicit the recurrsion is simply: \[b_{n+1}=1.1(b_n)+11\]

OpenStudy (amistre64):

where b0 = 100

OpenStudy (anonymous):

much clearer, and recursive due to n calling itself

OpenStudy (amistre64):

\[b_n = 100(1.1)^n + 11(\frac{1-1.1^{n-1}} {1-1.1})\] might be the equation to determine the nth year explicitly

OpenStudy (anonymous):

is this last definition considered recursive?

OpenStudy (amistre64):

definition never was my strong point :) i dont think it would be considered recursive, but it should get you the nth year by plugging in n=year. the bn+1 = 1.1(bn)+11; b0=100 is recursive since you cycle thru it n times but a google search would prolly help this to be more definitive

OpenStudy (anonymous):

I'm satisfied with the recursive definition, and laying out some of the initial steps makes it very evident...thank you!

OpenStudy (amistre64):

youre welcome, and good luck ;)

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