已知分别以d1和d2为公差的等差数列an和bn满足a1=18 ,b14=361,d1=18,且存在正整数m,使得am^2=b(m+14)-45,求证d2>1082,若ak=bk=0,a1,a2,...,ak,bk+1,bk+2,...,b14的前n项和Sn满足S14=2Sk,求an,bn通项3,在2的条件下,令Cn=a^a
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![已知分别以d1和d2为公差的等差数列an和bn满足a1=18 ,b14=361,d1=18,且存在正整数m,使得am^2=b(m+14)-45,求证d2>1082,若ak=bk=0,a1,a2,...,ak,bk+1,bk+2,...,b14的前n项和Sn满足S14=2Sk,求an,bn通项3,在2的条件下,令Cn=a^a](/uploads/image/z/1987728-24-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%88%86%E5%88%AB%E4%BB%A5d1%E5%92%8Cd2%E4%B8%BA%E5%85%AC%E5%B7%AE%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E5%92%8Cbn%E6%BB%A1%E8%B6%B3a1%3D18+%2Cb14%3D361%2Cd1%3D18%2C%E4%B8%94%E5%AD%98%E5%9C%A8%E6%AD%A3%E6%95%B4%E6%95%B0m%2C%E4%BD%BF%E5%BE%97am%5E2%3Db%28m%2B14%29-45%2C%E6%B1%82%E8%AF%81d2%EF%BC%9E1082%2C%E8%8B%A5ak%3Dbk%3D0%2Ca1%2Ca2%2C...%2Cak%2Cbk%2B1%2Cbk%2B2%2C...%2Cb14%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3S14%3D2Sk%2C%E6%B1%82an%2Cbn%E9%80%9A%E9%A1%B93%2C%E5%9C%A82%E7%9A%84%E6%9D%A1%E4%BB%B6%E4%B8%8B%2C%E4%BB%A4Cn%3Da%5Ea)
已知分别以d1和d2为公差的等差数列an和bn满足a1=18 ,b14=361,d1=18,且存在正整数m,使得am^2=b(m+14)-45,求证d2>1082,若ak=bk=0,a1,a2,...,ak,bk+1,bk+2,...,b14的前n项和Sn满足S14=2Sk,求an,bn通项3,在2的条件下,令Cn=a^a
已知分别以d1和d2为公差的等差数列an和bn满足a1=18 ,b14=36
1,d1=18,且存在正整数m,使得am^2=b(m+14)-45,求证d2>108
2,若ak=bk=0,a1,a2,...,ak,bk+1,bk+2,...,b14的前n项和Sn满足S14=2Sk,求an,bn通项
3,在2的条件下,令Cn=a^an,dn=a^bn,a>0,a≠1,问不等式CnDn+1≤Cn+Dn是否对正整数n恒成立?
已知分别以d1和d2为公差的等差数列an和bn满足a1=18 ,b14=361,d1=18,且存在正整数m,使得am^2=b(m+14)-45,求证d2>1082,若ak=bk=0,a1,a2,...,ak,bk+1,bk+2,...,b14的前n项和Sn满足S14=2Sk,求an,bn通项3,在2的条件下,令Cn=a^a
1.由an=a1+(n-1)*d,an=am+(n-m)d
得am=18+(m-1)*18=18m,b(m+14)=36+(m+14-14)*d2=36+(d2)m
得 am^2=b(m+14)-45
(18m)^2=36+(d2)m-45
d2=324m+(9/m)
因为m为正整数,所以 324m+(9/m)>108,所以d2>108
2.因为an和bn为等差数列,由Sn=n(a1+an)/2,S=(m-n+1)(an+am)/2
得 Sk=a1+a2+...+ak=k(a1+ak)/2=k(18+0)/2=9k
得 S14=a1+a2+...+ak+bk + bk+1 + bk+2 ...+ b14 - bk
S14=9k+{(bk+b14)*(14-k+1)/2}-bk
S14=9k+{(0+36)*(14-k+1)/2}-0
S14=9k+18*(15-k)
因为S14=2Sk
得 9k+18*(15-k)=2*9k
k=10
所以,a10=b10=0
所以,由an=a1+(n-1)*d,an=am+(n-m)d
a10=a1+(10-1)*d1,代入得d1=-2
b14=a10+(14-10)*d2,代入得d2=9
所以,通项an=20-2n,通项bn=9n-90