limit

find limit n tends to infinity ( 1/1.4 + 1/4.7+..........+1/ (3n-2) ( 3n+1)

2 Answers

This is the standard question which is simply

1/3 \left(1/4-1/7 + 1/7-1/11 + 1/11-1/14+...+1/(3n-2)-1/(3n+1) \right) = 1/3 (1/4-1/(3n+1))

T_n=1(3n-2)(3n+1)=13(13n-2 - 13n+1)

T_n+1=13(13n+1 - 13n+4)

similarly write T_n+2 , and then T_n+3..and so on and add dem up

actually problem is killed wen u realise that if a_n = 3n-2 then a_n+1 =3n+1

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