下面2題M1既數都唔識,
唔該幫幫手/_\
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唔該幫下手THX/_\
唔識2題M1數(急)
2010-12-16 4:59 am
回答 (3)
2010-12-16 5:45 am
✔ 最佳答案
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參考: me
2010-12-16 5:46 am
1(a) 3/(n-1)!+1/n!
=3n/n!+1/n!
=(3n+1)/n!
(b) 4/1!+7/2!+10/3!+13/4!+....
=[3/0!+1/1!]+[3/1!+1/2!]+[3/2!+1/3!]+[3/3!+1/4!]+...
=3[1/0!+1/1!+1/2!+1/3!+...]+[1/1!+1/2!+1/3!+1/4!+...]
=3e+(e-1)
=4e-1
2(a) (1-x^2)e^(2x)
=(1-x^2)[1+2x+(2x)^2/2!+(2x)^3/3!+...]
=(1-x^2)[1+2x+4x^2/2+8x^3/6+...]
=1+2x+4x^2/2+8x^3/6-x^2-2x^3
=1+2x+x^2-(2/3)x^3
(b) Comparing the coefficients, we have
a_0=1,a_1=2,a_2=1,a_3=-2/3
Generally,for n > 2, a_n=(2^n/n!)-2^(n-2)/(n-2)!
=[2^(n-2)(4-n(n-1)]/n!
=[2^(n-2)(4+n-n^2)]/n!
=3n/n!+1/n!
=(3n+1)/n!
(b) 4/1!+7/2!+10/3!+13/4!+....
=[3/0!+1/1!]+[3/1!+1/2!]+[3/2!+1/3!]+[3/3!+1/4!]+...
=3[1/0!+1/1!+1/2!+1/3!+...]+[1/1!+1/2!+1/3!+1/4!+...]
=3e+(e-1)
=4e-1
2(a) (1-x^2)e^(2x)
=(1-x^2)[1+2x+(2x)^2/2!+(2x)^3/3!+...]
=(1-x^2)[1+2x+4x^2/2+8x^3/6+...]
=1+2x+4x^2/2+8x^3/6-x^2-2x^3
=1+2x+x^2-(2/3)x^3
(b) Comparing the coefficients, we have
a_0=1,a_1=2,a_2=1,a_3=-2/3
Generally,for n > 2, a_n=(2^n/n!)-2^(n-2)/(n-2)!
=[2^(n-2)(4-n(n-1)]/n!
=[2^(n-2)(4+n-n^2)]/n!
2010-12-16 5:24 am
收錄日期: 2021-04-26 14:04:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101215000051KK01256