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的值为( )我是这样做的:令x=0,啧a0=n,an=1接下去呢?
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![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的值为( )我是这样做的:令x=0,啧a0=n,an=1接下去呢?](/uploads/image/z/2710107-27-7.jpg?t=%E5%B7%B2%E7%9F%A5%281+%2B+x+%29+%2B+%281+%2B+x+%292+%2B+%E2%80%A6+%2B+%281+%2B+x+%29n+%3D+a0+%2B+a1x+%2B+a2x2+%2B+%E2%80%A6+%2B+anxn%2C%E8%8B%A5a1+%2B+a2+%2B+a3+%2B+%E2%80%A6+%2B+an%EF%BC%8D1+%3D+29%EF%BC%8Dn%2C%E9%82%A3%E4%B9%88%E8%87%AA%E7%84%B6%E6%95%B0n%E7%9A%84%E5%80%BC%E4%B8%BA%EF%BC%88+%EF%BC%89%E6%88%91%E6%98%AF%E8%BF%99%E6%A0%B7%E5%81%9A%E7%9A%84%EF%BC%9A%E4%BB%A4x%3D0%2C%E5%95%A7a0%3Dn%2Can%3D1%E6%8E%A5%E4%B8%8B%E5%8E%BB%E5%91%A2%3F)
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的值为( )我是这样做的:令x=0,啧a0=n,an=1接下去呢?
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的值为( )
我是这样做的:令x=0,啧a0=n,an=1
接下去呢?
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的值为( )我是这样做的:令x=0,啧a0=n,an=1接下去呢?
令x=1
则1+x=2
所以左边是2+……+2^n=2^(n+1)-2
右边=a0+a1+……+a(n-1)+an=n+(29-n)+1=30
即2^(n+1)-2=30
2^(n+1)=32=2^5
n+1=5
n=4