Question

已知(1 + x ) + (1 + x )2 + … + (1 + x )n = a0 + a1x + a2x2 + … + anxn，若a1 + a2 + a3 + … + a

已知(1 + x ) + (1 + x )2 + … + (1 + x )n = a0 + a1x + a2x2 + … + anxn，若a1 + a2 + a3 + … + an－1 = 29－n，那么自然数n的值

Answer 1

在等式(1+x)+(1+x)²+······+(1+x)^n=a0+a1x+a2x+······+anxn中
令x=0得a0=n
令x=1得2+2²+······+2^n=a0+a1+a2+······+an
即a1+a2+······+an+n=2(2^n-1)
a1+a2+······+an=2(2^n-1)-n
又a1+a2+······+an-1=29-n，即a1+a2+······+an=30-n
所以2(2^n-1)-n=30-n
2^n-1=15
2^n=16
n=4