The given series is,3×8+6×11+9×14+...an=(nthterm of3,6,9...)×(nthterm of8,11,14...)=(3n)(3n+5)=9n2+15n∴Sn=∑nk=1ak=∑nk=1(9k2+15k)=9∑nk=1k2+15∑nk=1k=9×n(n+1)(2n+1)6+15×n(n+1)(2n+1)2=3n(n+1)(2n+1)2+15n(n+1)2=3n(n+1)2(2n+15)=3n(n+1)2(2n+6)=3n(n+1)(n+3)
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