Question

設(shè)n∈N^*，（2x+1）ⁿ的展開式各項(xiàng)系數(shù)之和為a_n，（3x+1）ⁿ展開式的二項(xiàng)式系數(shù)之和為b_n，則

lim

n→+∞

2an+3bn

an+1bn+1

=______．

Answer 1

令x=1由二項(xiàng)式定理可得a_n=3ⁿ，（3x+1）ⁿ展開式的二項(xiàng)式系數(shù)之和b_n=2ⁿ
∴

lim

n→∞

2an+3bn

an+1bn+1

=

lim

n→∞

2•3n+3•2n

3n+1+2n+1

=

lim

n→∞

2+3•( 2 3 )n

3+2•( 2 3 )n

=

2+3 lim n→∞ ( 2 3 )n

3+2 lim n→∞ ( 2 3 )n

=

2

3

故答案為

2

3