Functions, Can you solve this?

Algebra Level 5

f ( m + 1 ) = ( − 1 ) m + 1 m − 2 f ( m ) \large f(m+1) = (-1)^{m+1} m - 2f(m)

Let function f f , true for integers m m , satisfy the above equation and f ( 1 ) = f ( 2009 ) f(1) = f(2009) . If ∑ k = 1 2008 f ( k ) = Φ \displaystyle \sum_{k=1}^{2008} f(k) = \Phi , find 3 Φ 3 \Phi .


The answer is -1004.

Rishabh Tiwari by Rishabh Tiwari , 20, India

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1 solution

Moaz Mahmud
Apr 15, 2017

Φ = ∑ k = 1 2008 f ( k ) \Phi = \displaystyle \sum_{k=1}^{2008} f(k)

= ∑ k = 1 2008 { ( − 1 ) k ( k − 1 ) − 2 f ( k − 1 ) } = \displaystyle\sum\limits_{k = 1}^{2008} \{(-1)^{k} (k-1) - 2f(k-1)\}

= ∑ k = 1 2008 ( − 1 ) k ( k − 1 ) − 2 ∑ k = 1 2008 f ( k − 1 ) = \displaystyle\sum\limits_{k = 1}^{2008} (-1)^{k} (k-1) - 2\displaystyle\sum\limits_{k = 1}^{2008} f(k-1)

= ∑ k = 0 2007 ( − 1 ) k + 1 k − 2 ∑ k = 0 2007 f ( k ) = \displaystyle\sum\limits_{k = 0}^{2007} (-1)^{k+1}k - 2\displaystyle\sum\limits_{k = 0}^{2007} f(k)

= 1004 − 2 [ f ( 0 ) + ∑ k = 1 2008 f ( k ) − f ( 2008 ) ] = 1004 - 2[f(0)+\displaystyle\sum\limits_{k = 1}^{2008} f(k)-f(2008)]

= 1004 − 2 f ( 0 ) − 2 Φ + 2 f ( 2008 ) = 1004 - 2f(0)-2\Phi+2f(2008)

∴ 3 Φ = 1004 − 2 f ( 0 ) + 2 f ( 2008 ) \therefore 3\Phi = 1004 - 2f(0)+2f(2008)

Now, f ( 2009 ) = ( − 1 ) 2009 × 2008 − 2 f ( 2008 ) = f ( 1 ) = ( − 1 ) 1 × 0 − 2 f ( 0 ) f(2009) = (-1)^{2009}\times 2008 - 2f(2008) = f(1) = (-1)^{1}\times 0 - 2f(0)

∴ − 2 f ( 0 ) + 2 f ( 2008 ) = − 2008 \therefore -2f(0)+2f(2008) = -2008

∴ 3 Φ = 1004 − 2008 = − 1004 \therefore 3\Phi = 1004 - 2008 = -1004

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