Question

（2013•湛江一模）若(x+1)5=a0+a1(x-1)+a2(x-1)2+…+a5(x-1)5，則a₀=（　�。�

A．1

B．32

C．-1

D．-32

Answer 1

分析：根據(jù) （x+1）⁵=[2+（x-1）]⁵=

C

0 5

•2⁵+

C

1 5

•2⁴（x-1）+

C

2 5

•2³•（x-1）²+

C

3 5

•2²（x-1）³+

C

4 5

•2•（X-1）⁴+

C

5 5

•（x-1）⁵，結(jié)合所給的條件求得a₀的值．

解答：解：∵（x+1）⁵=[2+（x-1）]⁵=

C

0 5

•2⁵+

C

1 5

•2⁴（x-1）+

C

2 5

•2³•（x-1）²+

C

3 5

•2²（x-1）³+

C

4 5

•2•（X-1）⁴+

C

5 5

•（x-1）⁵，
而且 (x+1)5=a0+a1(x-1)+a2(x-1)2+…+a5(x-1)5，
故 a₀=

C

5 5

•2⁵=32，
故選B．

點(diǎn)評(píng)：本題主要考查二項(xiàng)式定理的應(yīng)用，二項(xiàng)式展開式的通項(xiàng)公式，求展開式中某項(xiàng)的系數(shù)，屬于中檔題．