Question

Find the sum of the following series to n terms 1.2.4+2.3.7+3.4.10+.....

Solution

Let Tnbe the nth term of this series. Then, Tn=(nthtermof1,2,3...)×(nthtermof2,3,4...)×(nthtermof4,7,10...)=[1+(n−1)×1].[2+(n−1)×1].[4+(n−1)×3]=[1+n−1].[2+n−1].[4+3n−3]=n(n+1)(3n+1)=(n2+n)(3n+1)=3n3+n2+3n2+n=3n3+4n2+n∴Tn=3n3+4n2+nLet Sn be the sum to n terms of the given series; Then,Sn=∑nn−1Tn=∑nn−1(3n3+4n2+n)=∑nn−13n3+∑nn−14n2+∑nn−1n=3∑nn−1n3+4∑nn−1n2+∑nn−1n=[n(n+1)2]2+4[n(n+1)(2n+1)2]+[n(n+1)2]=34[n(n+1)]2+2n(n+1)(2n+1)3+n(n+1)2=9[n(n+1)]2+8n(n+1)(2n+1)+6n(n+1)12=n(n+1)12[9n(n+1)+8(2n+1)+6]=n12(n+1)[9n2+9n+16n+8+6]=n12(n+1)[9n2+25n+14]Hence,Sn=n12(n+1)(9n2+25n+14)

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