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求证1+1/3^2+1/5^2+..+1/(2n%1)^2>7/6%1/2(2n+1)(n&g...

对于n≥5,即2n-1≥9,有:1+1/3^2+1/5^2+…+1/(2n-1)^2=(1+1/3^2+1/5^2+1/7^2)+1/9^2+1/11^2+…+1/(2n-1)^2,经计算1+1/3^2+1/5^2+1/7^2=1+1/9+1/25+1/49>1.17,因此,1+1/3^2+1/5^2+…+1/(2n-1)^2=(1+1/3^2+1/5^2+1/7^2)+1/9^2+1/11^2+

左边=1+1/(3*3)+1/(5*5)++1/[(2n-1)(2n-1)] =1+(1-1/3)/2+(1/3-1/5)/2++[1/(2n-3)-1/(2n-1)]/2 =1+1/2-1/[2(2n-1)] =3/2

证明:1+1/2^2+1/3^2++1/n^2<(2n-1)/n (n>=2,n属于n*) 1)1+1/2^2=5/4 < 3/2 2) 设:1+1/2^2+1/3^2++1/k^2<(2k-1)/k, 1+1/2^2+1/3^2++1/k^2+1/(k+1)^2<(2k-1)/k+1/(k+1)^2 =(2k^3+4k^2+2k-k^2-2k-1+k)/k(k+1)^2 =2-(k-1)/k(k+1)<[2(k+1)-1]/(k+1) 也就是如果n=k时成立能推出n=k+1也成立 所以,1+1/2^2+1/3^2++1/n^2<(2n-1)/n

令n=1,原不等式成立假设n=k时不等式成立.n=k+1时:把n=k时得到的不等式带入左边得:原式>7/6-1/2(2k+1)+1/(2k+1)^2=7/6-1/2(2k+3)+1/2(2k+3)-1/2(2k+1)+1/(2k+1)^2=7/6-1/2(2k+3)+[(2k+1)-(2k+3)/2(2k+3)(2k+1)]+1/(2k+1)^2=7/6-1/2(2k+3)-1/(2k+3)(2k+1)+1/(2k+1)^2=7/6-1/2(2k+3)-(2k+3)(2k+1)-(2k+1)^2/(2k+3)(2k+1)(2k+1)^2>7/6-1/2(2k+3)所以,对于任意正整数不等式恒成立

证明: 1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2= 1 / 2 * [ 1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2 + 1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2 ]> 1 / 2 * [ 1 + 1 / 3^2 + 1 / 5^2 + … + 1 / (2n-1)^2 + 1 + 1 / 4^2 + 1 / 6^2 + … + 1 / (2n)^2 ]= 1 / 2 * [ 2 + 1 / 3^2 + 1

1、当n=2时左边=1+1/3^2=10/9右边=7/6-1/2*3=1左边>右边不等式成立2、假设当n=K,K∈N,K>1时成立即1+1/3^2+1/5^2+..+1/(2K-1)^2>7/6-1/2(2K+1)那么当n=K+1时1+1/3^2+1/5^2+..+1/(2K-1)^2+1/(2K+1)^2>7/6-1/2(2K+1)+1/(2K+1)^2=7/6>7/6-1/2[2(K+1)+1]也成立所以1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1)得证【数学辅导团】为您解答,如果本题有什么不明白可以追问,

[4*6*8*……*(2n+2)]^2<(n+1)[3*5*7*9*……*(2n+1)]^2[4*6*8*……*(2n+2)]*[4*6*8*……*(2n+2)/(n-1)]<[3*5*7*9*……*(2n+1)]^2[4*6*8*……*(2n+2)]*[2*4*6*……*2n]<[3*5*7*9*……*(2n+1)]^2(4*2)*(6*4)*……*[(2n+2)*2n]<3^2*5^2*…… *2n+1)^2只需证明 (2n+2)*2n < (2n+1)^2即4n^2+4n<4n^2+4n+1

证明:当n=1时,左式=1^2,右式=1/3*(4-1)=1 左式=右式,等式成立 令 当n=k时,1^2+3^2+5^2+…+(2k-1)^2=1/3k*(4k^2-1) 成立 且k是大于等于2的正整数 那么 当n=k+1 时, 左边=1^2+3^2+5^2+…+(2k-1)^2+(2k+1)^2=1/3k*(4k^2-1)+(2k+1)^2

证明:n>1时1/n<1/(n-1)=1/[(n-1)(n+1)]=[1/(n-1)-1/(n+1)]/2所以左<1+1/4+(1/2)[1/2-1/4+1/3-1/5+1/4-1/6++1/(n-2)-1/n+1/(n-1)-1/(n+1)] =1+1/4+(1/2)[1/2+1/3-1/n-1/(n+1)] <5/4+(1/2)*(1/2+1/3) =5/4+(1/2)*(5/6) =5/4+5/12 =5/3得证.==========

证明:∵1+1/2^2+1/3^2+.+1/11^2+1/12^2+1/13^2= 1+1/4+1/9+.+1/11^2+1/12^2+1/ 12n+6∴n(n+1)>6(n+1)+6∴1/n+1/(n+1)5/3∴1+1/2{1-1/3+1/2-1/4+1/3-1/5+.+1/(n-4)-1/(n

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