两道高中数学题 急求解答
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(14)
(a1)^2+(a2)^2/2^2+...+(an)^2/n^2 = 4n-3 (1)
n=1, a1= 1
for n>=2
(a1)^2+(a2)^2/2^2+...+(a(n-1))^2/(n-1)^2 = 4(n-1)-3 (2)
(1)-(2)
(an)^2/n^2 = 4
an=2n
ie
an=1 ; n=1
=2n ; n=2,3,....
T20
=a1+(a2+a3+...+a20)
= 1+ (40+4)19/2
=1+418
=419
(15)
|a+b|=√3
|a|^2+|b|^2+2a.b=3 (1)
|a-b|=1
|a|^2+|b|^2-2a.b=1 (2)
(1)-(2)
4a.b= 2
a.b = 1/2
from (1)
|a|^2+|b|^2 = 2
|a||b| ≤ (|a|^2+|b|^2)/2 = 1
=>max |a||b| =1
(a1)^2+(a2)^2/2^2+...+(an)^2/n^2 = 4n-3 (1)
n=1, a1= 1
for n>=2
(a1)^2+(a2)^2/2^2+...+(a(n-1))^2/(n-1)^2 = 4(n-1)-3 (2)
(1)-(2)
(an)^2/n^2 = 4
an=2n
ie
an=1 ; n=1
=2n ; n=2,3,....
T20
=a1+(a2+a3+...+a20)
= 1+ (40+4)19/2
=1+418
=419
(15)
|a+b|=√3
|a|^2+|b|^2+2a.b=3 (1)
|a-b|=1
|a|^2+|b|^2-2a.b=1 (2)
(1)-(2)
4a.b= 2
a.b = 1/2
from (1)
|a|^2+|b|^2 = 2
|a||b| ≤ (|a|^2+|b|^2)/2 = 1
=>max |a||b| =1
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