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Find the sum of n terms :- 1 2 +4 2 +7 2 +............. Question based on Sequence and series.
its n terms will be (2n –1)2thus Tn = 4n^2 + 1 – 4nso Sn = TnSn = (4n^2 + 1 – 4n)now open sigma seperatelySn = 4n^2 + 1 – 4n put formula and you will get the answer Hope it helps
Its n term will be (3n-2)2 So the terms can be written as,9n2+12n+4,S Sn = 9(12+22+32+...+n2 )+12(1+2+3+...+n)+4= {9(n)(n+1)(2n+1)/6}+{12(n)(n+1)/2}+4
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