大学高数题,求数列极限。请好心人帮助一下

2025-01-03 05:10:35
推荐回答(2个)
回答1:

1²/n³+3²/n³+5²/n³+......+(2n-1)²/n³
=[1²+3²+5²+......+(2n-1)²]/n³
=[1²+2²+3²+......+(2n-1)²-2²-4²-6²-......-(2n-4)²]/n³
=[2n(2n-1)(4n-1)/6-4[1²-2²-3²-......-(n-2)²]/n³
=[2n(2n-1)(4n-1)/6-4(n-1)(n-2)(2n-3)/6]/n³
=[n(2n-1)(4n-1)/3-2(n-1)(n-2)(2n-3)/3]/n³
=(2n-1)[n(4n-1)-2(n-2)(2n-3)]/(3n³)
=(2n-1)[(4n²-n)-(2n²-7n+6)]/(3n³)
=(2n-1)[4n²-n-2n²+7n-6]/(3n³)
=(2n-1)[2n²+6n-6]/(3n³)
=(2-1/n)[2+6/n-6/n²]/3
lim(2-1/n)2+6/n-6/n²]/3
=2x2/3
=4/3

回答2:

数列极限是啥