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Question

If the nth term and the sum of n terms of the series 2,12,36,80,150,252,..... is Tn and Sn repectively then

A
Tn=n3+n2
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B
112(n)(n+1)(n+2)(3n+2)
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C
Tn=n3n2
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D
112(n)(n+1)(n+2)(3n+1)
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Solution

The correct options are
A Tn=n3+n2
D 112(n)(n+1)(n+2)(3n+1)
Let Sn=2+12+36+80+150+252+......+Tn(1)Sn=2+12+36+80+150+252+....+Tn1+Tn(2)On subtracting equation (2) from equation (1) ,We getTn=2+10+24+44+70+102+...+tn(3), where tn=TnTn1Tn= 2+10+24+44+70+102+...+tn1+tn(4)On subtracting equation (4) from equation (3) ,We gettn=2+8+14+20+26+...=n2[4+(n1)6]=n[3n1]tn=3n2nGeneral term of given series is Tn=tn=(3n2n)=n(n+1)(2n+1)2n(n+1)2=n3+n2Hence,sum of the series is Sn=n3+n2=n2(n+1)24+n(n+1)(2n+1)6=112(n)(n+1)(n+2)(3n+1)

Alternate Solution:
Substitute n=1,2,3 in the options given
and verify it with the given series

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