数列归纳法3/(1^2*2^2)+5/(2^2*3^2)+7/(3^2*4^2)+...+n^2(n+1)^2/2n+1=1-1/(n+1)^2
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数列归纳法3/(1^2*2^2)+5/(2^2*3^2)+7/(3^2*4^2)+...+n^2(n+1)^2/2n+1=1-1/(n+1)^2
数列归纳法3/(1^2*2^2)+5/(2^2*3^2)+7/(3^2*4^2)+...+n^2(n+1)^2/2n+1=1-1/(n+1)^2
数列归纳法3/(1^2*2^2)+5/(2^2*3^2)+7/(3^2*4^2)+...+n^2(n+1)^2/2n+1=1-1/(n+1)^2
证:
n=1时,3/(1²×2²)=3/4 1- 1/(1+1)²=1-1/4=3/4
3/(1²×2²)=1- 1/(1+1)²,等式成立.
假设当n=k(k∈N+,k≥1)时,等式成立,即
3/(1²×2²)+5/(2²×3²)+7/(3²×4²)+...+(2k+1)/[k²(k+1)²]=1- 1/(k+1)²
则当n=k+1时,
3/(1²×2²)+5/(2²×3²)+7/(3²×4²)+...+(2k+1)/[k²(k+1)²]+[2(k+1)+1]/[(k+1)²(k+1+1)²]
=1- 1/(k+1)² +(2k+3)/[(k+1)²(k+2)²]
=1-[(k+2)²-(2k+3)]/[(k+1)²(k+2)²]
=1-(k²+4k+4-2k-3)/[(k+1)²(k+2)²]
=1-(k²+2k+1)/[(k+1)²(k+2)²]
=1- (k+1)²/[(k+1)²(k+1+1)²]
=1- 1/[(k+1)+1]²,等式同样成立.
k为任意正整数,因此等式对于任意正整数n恒成立.
3/(1²×2²)+5/(2²×3²)+7/(3²×4²)+...+(2n+1)/[n²(n+1)²]=1- 1/(n+1)²