已知等差数列(an)的前n项和为Sn,且满足;a3=7,a5+a7=26
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已知等差数列(an)的前n项和为Sn,且满足;a3=7,a5+a7=26
若m=(2^an)/(2^n+2),数列(bn)满足关系式bn=(1 (n=1)
b(n-1)+m (n>1)
求数列(bn)的通项公式
若m=(2^an)/(2^n+2),数列(bn)满足关系式bn=(1 (n=1)
b(n-1)+m (n>1)
求数列(bn)的通项公式
a5+a7=2a6=26
a6=13
a6=a3+3d
13=7+3d
3d=6
d=2
a3=a1+2d
7=a1+2*2
a1=3
an=a1+(n-1)d
=3+2(n-1)
=2n+1
m=(2^an)/2^(n+2)
=2^(2n+1)/2^(n+2)
=2^(2n+1-n-2)
=2^(n-1)
bn=b(n-1)+m
bn-b(n-1)=m
bn-b(n-1)=2^(n-1)
.
b3-b2=2^2
b2-b1=2^1
以上等式相加得
bn-b1=2^1+2^2+.+2^(n-1)
bn-b1=2*[1-2^(n-1)]/(1-2)
bn-1=2*[2^(n-1)-1]
bn-1=2^n-2
bn=2^n-1
a6=13
a6=a3+3d
13=7+3d
3d=6
d=2
a3=a1+2d
7=a1+2*2
a1=3
an=a1+(n-1)d
=3+2(n-1)
=2n+1
m=(2^an)/2^(n+2)
=2^(2n+1)/2^(n+2)
=2^(2n+1-n-2)
=2^(n-1)
bn=b(n-1)+m
bn-b(n-1)=m
bn-b(n-1)=2^(n-1)
.
b3-b2=2^2
b2-b1=2^1
以上等式相加得
bn-b1=2^1+2^2+.+2^(n-1)
bn-b1=2*[1-2^(n-1)]/(1-2)
bn-1=2*[2^(n-1)-1]
bn-1=2^n-2
bn=2^n-1
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