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Question

Find the general term and the sum of n terms of the series 12,40,90,168,280,432,.....

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Solution

The successive orders of difference are 28,50,78,112,152,....
22,28,34,40,....
6,6,6,....
0,0,...
Hence the nth term =12+28(n1)+22(n1)(n2)2+6(n1)(n2)(n3)3
=n3+5n2+6n.
The sum of n terms may now be found by writing down the value of n3+5n2+6n. Or we may use the formula of the present article and obtain Sn=12n+28n(n1)2+22n(n1)(n2)3+6n(n1)(n2)(n3)4
=n12(3n2+26n+69n+46),
=112n(n+1)(3n2+23n+46).

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