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5.2.13-eg - k starting with k = 0 so the entries in the sum...

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Example (a) Find a function f ( k ) so that 3 + 6 + 9 + 12 = 4 X k =1 f ( k ) . (b) Find a function g ( k ) so that 3 + 6 + 9 + 12 = 3 X k =0 g ( k ) . Solution: (a) For the first summation the index of summation is k , starting with k = 1, the entries in the sum can be identified as follows 3 |{z} k =1 + 6 |{z} k =2 + 9 |{z} k =3 + 12 |{z} k =4 , so the formula for the k th term can be found by noting that 3 = 3 · k, for k = 1 , 6 = 3 · k, for k = 2 , 9 = 3 · k, for k = 3 , 12 = 3 · k, for k = 4 . Thus, f ( k ) = 3 k. (b) Here the index of summation is k , starting with
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Unformatted text preview: k , starting with k = 0, so the entries in the sum can be identified as follows 3 |{z} k =0 + 6 |{z} k =1 + 9 |{z} k =2 + 12 |{z} k =3 . Writing each term as a function of k , 3 = 3 · 1 = 3 · ( k + 1) , for k = 0 , 6 = 3 · 2 = 3 · ( k + 1) , for k = 1 , 9 = 3 · 3 = 3 · ( k + 1) , for k = 2 , 12 = 3 · 4 = 3 · ( k + 1) , for k = 3 , Thus, g ( k ) = 3( k + 1) ....
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