If (12−t1)+(22−t2)+.......+(n2−tn)=13n(n2−1) then tn is n .
A
True
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B
False
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Solution
The correct option is A True ∑n2−Sn=13n(n2−1) ∴Sn=16n(n+1)(2n+1)−13n(n+1)(n−1) Or ∴Sn=16n(n+1).(3)=12n(n+1)=∑n = 1 + 2 + 3......n ∴tn=n Alt.tn=Sn−Sn−1=12[n(n+1)(n−1)n]=12n.2=n