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Question

Consider a sequence whose sum to n terms is given by the quadratic function Sn=3n2+5n then sum of the squares of the first three terms of the given series is,

A
1100
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B
660
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C
799
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D
1000
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Solution

The correct option is B 660
Given: Sn=3n2+5n

Tn=SnSn1

Tn=(3n2+5n)(3(n1)2+5(n1))

Tn=(3n2+5n)(3(n22n+1))5n+5

Tn=3n2+5n3n2+6n35n+5

Tn=6n+2

Again T1=6×1+2=8,T21=82=64

T2=6×2+2=14,T22=142=196

T3=6×3+2=20,T23=202=400

T21+T22+T23=64+196+400=660

Hence option B is the answer

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