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A problem in Tianjin liberal arts mathematics college entrance examination
a 1=0,a2=2,a3 = 4; A4=8, a5= 12, a6= 18 (available according to conditions because k is a natural number) a7=24, a8=32.

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