The sum of n terms of an AP is 4n(n-1),then what is the sum of their squares

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Please find below the solution to the asked query:

We haveSn=4nn-1If Tn denotes nth term, thenT1=S1T1=4×11-1T1=0S2=T1+T24×22-1=0+T2T2=8Hence A.P. is 0,8,16,.....Tn=0+n-18Tn=8n-1We needS=n=1n Tn2=64n=1n n-12=64n=1n n2-2n+1=64 n=1nn2-2n=1nn+n=1n1=64nn+12n+16-2nn+12+n=64nn+12n+16-nn+1+n

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