If tn=1/4 (n+2)(n+3) for n=1,2,3....then find the value of 1/t1 + 1/t2 + 1/t3+....+t1/2003
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Answer:
4006 / 3009
Step-by-step explanation:
tn = (n+2)(n+3) / 4
1 / tn = 4 / (n+2)(n+3)
1/t1 + 1/t2 + 1/t3+.....+ 1/t2003
> 4 [ 1/3.4 + 1/4.5 + 1/5.6 + .... + 1/2005.2006]
> 4 [(1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6) + .... + (1/2005 - 1/2006)]
> 4 [1/3 - 1/2006]
> 4 [2006-3 / 3.2006]
> 4 * 2003 / 3 * 2006
> 4006 / 3009
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