5.2.3-eg-1

# 5.2.3-eg-1 - -1 k =-1 for k odd 1 for k even so-1 Â-1 k =-1...

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Example Find a function f ( k ) so that 3 - 6 + 9 - 12 + 15 - 21 + 24 = 7 X k =1 f ( k ) . Solution: Without the alternating signs on the terms, it is not diﬃcult to see that 3 + 6 + 9 + 12 + 15 + 21 + 24 = 7 X k =1 (3 k ) . Now, including the alternating signs, 3 |{z} k =1 + ( - 6) | {z } k =2 + 9 |{z} k =3 + ( - 12) | {z } k =4 + 15 |{z} k =5 + ( - 21) | {z } k =6 + 24 |{z} k =7 , so the even terms are negative and the odd terms are positive. But
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Unformatted text preview: (-1) k = (-1 , for k odd , 1 , for k even so (-1) Â· (-1) k = (-1) k +1 = (-(-1) = 1 , for k odd ,-1 , for k even gives the correct alternating pattern of negative/positive entries. It follows that 3-6 + 9-12 + 15-21 + 24 = 7 X k =1 (-1) k +1 (3 k ) , so that f ( k ) = (-1) k +1 (3 k ) ....
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