Let a_1=1 and for n ⩾1 ,a_n+1= 12a_n+ n^2-2n-1n^2 (n+1 )^2. Then ( 0.05em 0.05em(a_n- 2n^2)) is equal to \_ \_ \_ \_ .

Question

Accepted Answer

Correct answer: 2. (2) a_n+1= 12a_n+ 2n^2- (n+1 )^2n^2 (n+1 )^2 a_n+1= 12a_n+ 2 (n+1 )^2- 1n^2 (a_n+1- 2 (n+1 )^2)= 12(a_n- 2n^2) T_n+1= 12T_n T_1 ,T_2 ,T_3 … in GP n=1T_1= 12(a_1- 21)=-1 ( _n=1^ 0.05em(a_n-2/n^2) )=(T_1+T_2+T_3 … . ) = ( -11- 12 )=2

$Let a_{1} = 1 and for n ⩾ 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}} . Then (\sum (a_{n} - \frac{2}{n ^{2}})) is equal to____.$

Solution

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