a 1=0*2+0*2
a2=0*2+0*2+ 1*2
a3=0*2+0*2+ 1*2+ 1*2
a4 = 0 * 2+0 * 2+ 1 * 2+ 1 * 2+2 * 2
a5 = 0 * 2+0 * 2+ 1 * 2+ 1 * 2+2 * 2+2 * 2
a6 = 0 * 2+0 * 2+ 1 * 2+ 1 * 2+2 * 2+2 * 2+3 * 2
a(2 * k- 1)= 0 * 2+0 * 2+ 1 * 2+ 1 * 2+2 * 2+2 * 2+3 * 2+.....+(k- 1)*2+(k- 1)*2
a(2 * k)= 0 * 2+0 * 2+ 1 * 2+ 1 * 2+2 * 2+2 * 2+3 * 2+.....+(k- 1)* 2+(k- 1)* 2+k * 2
a(2*k- 1)=2*k*(k- 1)
a(2*k)=2*k^2
Inductive verification
3.Tn'= Tn, but let an = n 2/2, where n is an arbitrary natural number. Let's eat
So: Tn'=2*n-2
Similarly: Tn''=Tn. But an = (n 2-1)/2 takes n=2*k- 1.
Therefore: TN'' = sum (n 2/(n 2-1)) n = 2345. .....
tn''=2*n-sum( 1/(n^2- 1))<； 2 * n-sum (1/n 2) = 2 * n-eee is a natural logarithm.
The inequality thus obtained. When n=2, the equal sign holds.
sum( 1/n^2)=e