20,若干个数,a1,a2,a3,...,an,若a1=2分之-1,从第二个数起,每个数都等于1与它前面的那个数的差的倒数.(1)a2= ;a3= ;a6= ; (2)求a9乘以a10乘以a11、an的值; (3)是否存在数M,使M除以(an-1乘以an乘
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![20,若干个数,a1,a2,a3,...,an,若a1=2分之-1,从第二个数起,每个数都等于1与它前面的那个数的差的倒数.(1)a2= ;a3= ;a6= ; (2)求a9乘以a10乘以a11、an的值; (3)是否存在数M,使M除以(an-1乘以an乘](/uploads/image/z/1562726-38-6.jpg?t=20%2C%E8%8B%A5%E5%B9%B2%E4%B8%AA%E6%95%B0%2Ca1%2Ca2%2Ca3%2C...%2Can%2C%E8%8B%A5a1%3D2%E5%88%86%E4%B9%8B-1%2C%E4%BB%8E%E7%AC%AC%E4%BA%8C%E4%B8%AA%E6%95%B0%E8%B5%B7%2C%E6%AF%8F%E4%B8%AA%E6%95%B0%E9%83%BD%E7%AD%89%E4%BA%8E1%E4%B8%8E%E5%AE%83%E5%89%8D%E9%9D%A2%E7%9A%84%E9%82%A3%E4%B8%AA%E6%95%B0%E7%9A%84%E5%B7%AE%E7%9A%84%E5%80%92%E6%95%B0.%EF%BC%881%EF%BC%89a2%3D+%EF%BC%9Ba3%3D+%EF%BC%9Ba6%3D+%EF%BC%9B+%282%29%E6%B1%82a9%E4%B9%98%E4%BB%A5a10%E4%B9%98%E4%BB%A5a11%E3%80%81an%E7%9A%84%E5%80%BC%EF%BC%9B+%EF%BC%883%EF%BC%89%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E6%95%B0M%2C%E4%BD%BFM%E9%99%A4%E4%BB%A5%EF%BC%88an-1%E4%B9%98%E4%BB%A5an%E4%B9%98)
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20,若干个数,a1,a2,a3,...,an,若a1=2分之-1,从第二个数起,每个数都等于1与它前面的那个数的差的倒数.(1)a2= ;a3= ;a6= ; (2)求a9乘以a10乘以a11、an的值; (3)是否存在数M,使M除以(an-1乘以an乘
20,若干个数,a1,a2,a3,...,an,若a1=2分之-1,从第二个数起,每个数都等于1与它前面的那个数的
差的倒数.(1)a2= ;a3= ;a6= ; (2)求a9乘以a10乘以a11、an的值; (3)是否存在数M,使M除以(an-1乘以an乘以an+1)=a1?若存在,请求出数M
20,若干个数,a1,a2,a3,...,an,若a1=2分之-1,从第二个数起,每个数都等于1与它前面的那个数的差的倒数.(1)a2= ;a3= ;a6= ; (2)求a9乘以a10乘以a11、an的值; (3)是否存在数M,使M除以(an-1乘以an乘
a1=-1/2,a2=2/3,a3=3,(a4=-1/2,a5=2/3) a6=3
我们发现该数列周期为3,所以a9*a11*a12=a3*a1*a2=-1,
-1/2 当n=3k+1,k∈N
an= 2/3 当n=3k+2,k∈N
3 当n=3k,k∈N
3. 因为an-1*an*an+1=-1,所以只需要M/-1=a1=-1/2.即M=1/2