Question

4．已知x⁵（x+3）³=a₈（x+1）⁸+a₇（x+1）⁷+…+a₁（x+1）+a₀，則7a₇+5a₅+3a₃+a₁=（　�。�

• A． -16
• B． -8
• C． 8
• D． 16

Answer 1

分析 根據(jù)x⁵（x+3）³ =[（x+1）-1]⁵•[（x+1）+2]³，按照二項式定理展開，求得a₇、a₅、a₃、a₁的值，可得7a₇+5a₅+3a₃+
a₁的值．

解答 解：∵x⁵（x+3）³=a₈（x+1）⁸+a₇（x+1）⁷+…+a₁（x+1）+a₀ =[（x+1）-1]⁵•[（x+1）+2]³ 
=[${C}_{5}^{0}$•（x+1）⁵-${C}_{5}^{1}$•（x-1）⁴+${C}_{5}^{2}$•（x-1）³-${C}_{5}^{3}$•（x-1）²+${C}_{5}^{4}$•（x-1）-${C}_{5}^{5}$]
•[${C}_{3}^{0}$•（x+1）³+2${C}_{3}^{1}$•（x+1）²+4${C}_{3}^{2}$•（x+1）+8${C}_{3}^{3}$]，
∴a₇ =${C}_{5}^{0}$•2${C}_{3}^{1}$-${C}_{5}^{1}$•${C}_{3}^{0}$=6-5=1，a₅=${C}_{5}^{0}$•8${C}_{3}^{3}$-${C}_{5}^{1}$•4${C}_{3}^{2}$+${C}_{5}^{2}$•2${C}_{3}^{1}$-${C}_{5}^{3}$•${C}_{3}^{0}$=8-60+60-10=-2，
a₃ =${C}_{5}^{2}$•8${C}_{3}^{3}$-${C}_{5}^{3}$•4${C}_{3}^{2}$+${C}_{5}^{4}$•2${C}_{3}^{1}$-${C}_{5}^{5}$•${C}_{3}^{0}$=80-120+30-1=-11，a₁=${C}_{5}^{4}$•8${C}_{3}^{3}$-${C}_{5}^{5}$•4${C}_{3}^{2}$=40-12=28，
∴7a₇+5a₅+3a₃+a₁=7-10-33+28=-8，
故選：B．

點評 本題主要考查二項式定理的應(yīng)用，二項展開式的通項公式，二項式系數(shù)的性質(zhì)，屬于中檔題．