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please help me by solving this question. i am unable to do so...
we know that std results: 1^2+2^2+----+r^2= r(r+1)(2r+1)/6 and 1^3+2^3+----+r^3= [r(r+1)/2]^2so, tr= 2(2r+1)/[3r(r+1)]now, note that we can write 3tr/2= (1/r) + 1/(r+1), since (1/r) + 1/(r+1)= (r+1+r)/r(r+1)= (2r+1)/r(r+1)so, (-1)^r3tr/2= (-1)^r/r – (-1)^(r+1)/(r+1) bcoz (-1)^(r+1)= – (-1)^rso, Sn= ∑(-1)^r*tr= (2/3)*∑(-1)^r/r – (-1)^(r+1)/(r+1) now, note that if we let f(r)= (-1)^r/r, then (-1)^(r+1)/(r+1)= f(r+1), and hence the series becomes telescopic.so, 3Sn/2= ∑f(r) – f(r+1)= f(1) – f(2)+f(2) – f(3)+f(3) – f(4)+.............+f(n) – f(n+1)= f(1) – f(n+1)or Sn= (2/3)*((-1)^1/1 – (-1)^(n+1)/(n+1)).put lim n tends to infS(inf) = – 2/3 so, lim n tends to inf (1/3 – Sn)= 1/3 – ( – 2/3)= 1/3 + 2/3= 1KINDLY APPROVE :))
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