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If tn=1/4(n+2)(n+3) for n=1,2,3,... then 1/t1+1/t2+1/t3+........+1/2003 |
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Answer» Tn=1/4(n+2)(n+3) =>1/Tn=4/((n+2)(n+3)) =>1/Tn=4[1/(n+2)-1/(n+3) Putting n=1,2,3,4......up to n and summing we get 1/T1+1/T2+1/T3+...Tn=.4(1/3-1/(n+3)) As n-> infinity We get 1/T1+1/T2+1/T3+.....=4/3 So 1/T1+1/T2+1/T3+.......+.1/2003 =4/3+1/2003 =8015/6009 |
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