Question

数学问题!!!!!!!急急!!!!!好的给分!!!!!

(√1)^2+1=2,S1=√1/2,
(√2)^2+1=3,S2=√2/2
(√3)^2+1=4,S3=√3/2,
请用n的等式表示上述变化规律
推算出OA10的长
求出S1^2+S2^2.......+S10^2的值

Answer 1

(√n)^2+1=n+1
Sn=√n/2

OA10=√10
S1^2+S2^2.......+S10^2
=1+2+3+4+5+6+7+8+9+10
=55

Answer 2

y=dx/2(√（49-x2）)+d49/2arcsin(x/7)
=1/2d√(49-x2)+√(49-x2)d(x/2)+49/2*1/√[1-x2/49)]d(x/7)
=1/2*1/[2√(49-x2)]*(-2x)dx+√(49-x2)/2dx+49/[2√(49-x2)]dx
={-x/[2√(49-x2)]+√(49-x2)/2+49/[2√(49-x2)]}dx