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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 Fall '10
 KIHYUNHYUN
 Calculus, ﬁrst summation

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