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=n3−n2
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D
112(n)(n+1)(n+2)(3n+1)
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Solution
The correct options are ATn=n3+n2 D112(n)(n+1)(n+2)(3n+1) Let Sn=2+12+36+80+150+252+......+Tn⋯(1)Sn=2+12+36+80+150+252+....+Tn−1+Tn⋯(2)On subtracting equation (2) from equation (1) ,We getTn=2+10+24+44+70+102+...+tn⋯(3),where tn=Tn−Tn−1Tn=2+10+24+44+70+102+...+tn−1+tn⋯(4)On subtracting equation (4) from equation (3) ,We gettn=2+8+14+20+26+...=n2[4+(n−1)6]=n[3n−1]tn=3n2−n∴General term of given series is Tn=∑tn=∑(3n2−n)=n(n+1)(2n+1)2−n(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