But we do not know
(a) h(n)
(b) h(n+1)
(c) h(n+1) - h(n)
We know h(x) = 2x^2 + x
2007-08-22 6:13 pm
回答 (3)
2007-08-22 7:07 pm
✔ 最佳答案
好簡單姐 你好似代數咁代入d式到(a)h(n) =2n^2+n//
(b)h(n+1)=2(n+1)^2+n+1
=2(n^2+2n+1)+n+1
=2n^2+4n+2+n+1
=2n^2+5n+3//
(c)h(n+1)-h(n)=(b)-(a)
=2n^2+5n+3-2n^2-n
=4n+3//
2007-08-22 7:01 pm
(a) h(x) =2x^2+x
h(n) = 2n^2+n
(b) h(x) = 2x^2+x
h(n+1) = 2(n+1)^2+ (n+1)
(c) h(x) = 2x^2+x
h(n+1) - h(n) = 2(n+1)^2+ (n+1) -2n^2+n
h(n) = 2n^2+n
(b) h(x) = 2x^2+x
h(n+1) = 2(n+1)^2+ (n+1)
(c) h(x) = 2x^2+x
h(n+1) - h(n) = 2(n+1)^2+ (n+1) -2n^2+n
2007-08-22 6:20 pm
a) h(x) =2x^2+x
h(n) = 2n^2+n
b)h(x) = 2x^2+x
h(n+1) = 2(n+1)^2+ (n+1)
c)h(x) = 2x^2+x
h(n+1) - h(n) = 2(n+1)^2+ (n+1) -2n^2+n
h(n) = 2n^2+n
b)h(x) = 2x^2+x
h(n+1) = 2(n+1)^2+ (n+1)
c)h(x) = 2x^2+x
h(n+1) - h(n) = 2(n+1)^2+ (n+1) -2n^2+n
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