Question

用数学归纳法证明3^2+5^2+.+(2n+1)^2=n/3()4n^+12n+11)

Answer 1

n=1,

3^2=1/3(4+12+11)成立

假设n=k时,

3^2+...(2k+1)^2=k/3(4k^2+12k+11)

则n=k+1时

左边

=3^2+...+(2k+1)^2+(2k+3)^2

=k/3(4k^2+12k+11)+(2k+3)^2

=1/3(4k^3+24k^2+47k+27)

=(k+1)/3(4k^2+20k+27)

=右边

故成立

=(k+1)/3(4(k+1)^2+12(k+1)+11)