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.
i always find it helpful to write out a few "terms" of the sequence to see if a pattern develops
b1 = (100*.1) + 11 b2 = b1*.1 + 11 = ((100*.1 +11) *.1) + 11 etc
1.1 is a better fit
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
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
It looks right, does this satisfy the definition of recursiveness?
Being that the last part is a geometric series, would it be better to write it in sigma summation notation?
im thinking this is turning into explicit the recurrsion is simply: \[b_{n+1}=1.1(b_n)+11\]
where b0 = 100
much clearer, and recursive due to n calling itself
\[b_n = 100(1.1)^n + 11(\frac{1-1.1^{n-1}} {1-1.1})\] might be the equation to determine the nth year explicitly
is this last definition considered recursive?
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
I'm satisfied with the recursive definition, and laying out some of the initial steps makes it very evident...thank you!
youre welcome, and good luck ;)
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