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For each positive integer n , let f(n+1)...

For each positive integer `n ,` let `f(n+1)=n(-1)^(n+1)-2f(n)` and `f(1)=(2010)dot` Then `sum_(k=1)^(2009)f(K)` is equal to 335 (b) 336 (c) 331 (d) 333

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For each positive integer n, let f(n+1)=n(-1)^(n+1)-2f(n) and f(1)=(2010)* Then sum_(k=1)^(2009)f(K) is equal to 335(b)336(c)331(d)333

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