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B
2n2+2n
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C
3n2+n
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D
2n2+2
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Solution
The correct option is C3n2+n Let the nth term be an2+bn+c, then First term =a(1)2+b(1)+c=a+b+c=4 Second term =a(2)2+b(2)+c=4a+2b+c=14 Third term =a(3)2+b(3)+c=9a+3b+c=30 Solving these three equations, we get a=3,b=1,c=0 Therefore, nth term =3n2+n