数列求和Sn=1/1*3+4/3*5+9/5*7+……n^2/(2n-1)(2n+1)

x1=2cosa
y1=4sina
设那点是Q
则A(2cosa+4sina,2cosa-4sina)
x=2cosa+4sina
y=2cosa-4sina
所以x+y=4cosa
x-y=8sina
sin²a+cos²a=1
所以(x+y)²/16+(x-y)²/64=1
4x²+8xy+4y²+x²-2xy+y²=64
5x²+6xy+5y²-64=0

n²/(2n-1)(2n+1)
=(n²-1/4+1/4)/(2n-1)(2n+1)
=(n²-1/4)/(2n-1)(2n+1)+(1/4)/(2n-1)(2n+1)
=(n+1/2)(n-/2)/(2n-1)(2n+1)+(1/8)[2/(2n-1)(2n+1)]
=(n+1/2)(n-/2)/[4(n+1/2)(n-/2)]+(1/8)[(2n+1)-(2n-1)]/(2n-1)(2n+1)]
=1/4+(1/8)[(2n+1)/(2n-1)(2n+1)-(2n-1)/(2n-1)(2n+1)]
=1/4+(1/8)[1/(2n-1)-1/(2n+1)]
所以原式=1/4+1/4+……+1/4+(1/8)[1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)]
=n/4+(1/8)[1-1/(2n+1)]
=n/4+n/(8n+4)
=(n²+n)/(4n+2)