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Question

Find the general term and the sum of n terms of the series 1,0,1,8,29,80,193,....

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Solution

10182980193....

1172151113....

26143062....

481632....

un=a.2n1+bn2+cn+d

Putting n=1,2,3,4 we get;-

a=4,b=1,c=2,d=0

un=4.2n1n22n+0

=2n+1n22n

Sn=un

=(22+23+....+2n+1)n22n

=4(2n1)16n(n+1)(2n+1)n(n+1)

=2n+2416n(n+1)(2n+7)


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