设数列{an},{bn}是分别以d1,d2为公差的等差数列,a1=50,b51
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设数列{an},{bn}是分别以d1,d2为公差的等差数列,a1=50,b51
设数列{an},{bn}是分别以d1,d2为公差的等差数列,a1=50,b51=100.(1) 若ak=bk=0,且数列a1,a2,...,ak,bk+1,bk+2,...,b51的前n项和为Sn,若S51=3Sk,求数列{an},{bn}通项公式.(2)在符合(1)的条件下,求数列{(-d1)^(bn/4)+26}的前n项和Tn
设数列{an},{bn}是分别以d1,d2为公差的等差数列,a1=50,b51=100.(1) 若ak=bk=0,且数列a1,a2,...,ak,bk+1,bk+2,...,b51的前n项和为Sn,若S51=3Sk,求数列{an},{bn}通项公式.(2)在符合(1)的条件下,求数列{(-d1)^(bn/4)+26}的前n项和Tn
先将两个数列的一部分列排列如下:
a1,a2,a3,.,ak
bk,bk+1,.b51
S51=(a1+a2+.+ak)+(b[k+1]+b[k+2]+.+b51)
=(50+0)*k/2+(bk+b[k+1]+b[k+2]+.+b51) (因为bk=0,加入后对和的大小没影响)
=25k+(0+100)*(52-k)/2
=25k+2600-50k
=2600-25k
又因为S51=3Sk=3*(50+0)*k/2=75k
所以
75k=2600-25k
k=26
d1=(a26-a1)/25=(0-50)/25=-2
d2=(b51-b26)/25=(100-0)/25=4
b1=b26-25d2=0-25*4=-100
an=a1+(n-1)d1=50+(n-1)*(-2)=52-2n
bn=b1+(n-1)d2=-100+(n-1)*4=-104+4n
--------------------------------------------------------
设数列{cn}={(-d1)^(bn/4)+26}
cn=2^[(-104+4n)/4]+26
=2^(n-26)+26
=(2^n)/(2^26)+26
则Tn=c1+c2+.+cn
=[2/(2^26)+26]+[4/(2^26)+26]+[8/(2^26)+26]+.+[(2^n)/(2^26)+26]
=[2*(2^n-1)/(2-1)]/(2^26)+26n
=[2^(n+1)-2]/(2^26)+26n
=2^(n-25)-2^(-25)+26n
a1,a2,a3,.,ak
bk,bk+1,.b51
S51=(a1+a2+.+ak)+(b[k+1]+b[k+2]+.+b51)
=(50+0)*k/2+(bk+b[k+1]+b[k+2]+.+b51) (因为bk=0,加入后对和的大小没影响)
=25k+(0+100)*(52-k)/2
=25k+2600-50k
=2600-25k
又因为S51=3Sk=3*(50+0)*k/2=75k
所以
75k=2600-25k
k=26
d1=(a26-a1)/25=(0-50)/25=-2
d2=(b51-b26)/25=(100-0)/25=4
b1=b26-25d2=0-25*4=-100
an=a1+(n-1)d1=50+(n-1)*(-2)=52-2n
bn=b1+(n-1)d2=-100+(n-1)*4=-104+4n
--------------------------------------------------------
设数列{cn}={(-d1)^(bn/4)+26}
cn=2^[(-104+4n)/4]+26
=2^(n-26)+26
=(2^n)/(2^26)+26
则Tn=c1+c2+.+cn
=[2/(2^26)+26]+[4/(2^26)+26]+[8/(2^26)+26]+.+[(2^n)/(2^26)+26]
=[2*(2^n-1)/(2-1)]/(2^26)+26n
=[2^(n+1)-2]/(2^26)+26n
=2^(n-25)-2^(-25)+26n
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