recursion

a_n = \left(1 - \frac{1}{n^2} \right)a_{n-1}-\frac{1}{n^2} \Rightarrow a_n+1 = (a_{n-1}+1) \left(1 - \frac{1}{n^2} \right)

Let b_n = a_n+1

Then b_n = b_{n-1} \left(1 - \frac{1}{n^2} \right)

Hence, \frac{b_{2000}}{b_1} = \frac{b_{2000}}{b_{1999}} \times \frac{b_{1999}}{b_{1998}}\times...\times \frac{b_{2}}{b_{1}}

=\prod_{k=2}^{2000} \frac{k^2-1}{k^2} = \prod_{k=2}^{2000} \frac{k-1}{k} \prod_{k=2}^{2000} \frac{k+1}{k} = \frac{1}{2000} \times \frac{2001}{2} = \frac{2001}{4000}

Hence b_{2000} = \frac{2001}{4000} and hence

a_{2000} = \frac{2001}{4000}-1 = -\frac{1999}{4000}

sir i have asked a2010,

but the solution is correct for a2000

great solution

OMG! About time i got myself checked up i think.