If tn = (1/4) (n+2)(n+3) for n = 1,2,3... then (1/t1 ) + (1/t2) + (1/t3) +....+ (1/t2005) =

Question

If tn = (1/4) (n+2)(n+3) for n = 1,2,3... then (1/t1 ) + (1/t2) + (1/t3) +....+ (1/t2005) =  (A) (4004/3005) (B) (3005/4004) (C) (8020/6024) (D) (8008/6015)

Tardigrade Admin · Accepted Answer

(1/tn) = (4/(n+2)(n+3)) = (4/n+2)-(4/n+3) Now ((4/3) - (4/4))+((4/4)-(4/5))+((4/5)- (4/6)) +.... + ((4/2007) - (4/2008)) = (4/3) - (4/2008)= (8020/6024)

Q. If $t_{n} = \frac{1}{4} (n + 2) (n + 3)$ for n = 1,2,3... then $\frac{1}{t _{1}} + \frac{1}{t _{2}} + \frac{1}{t _{3}} + .... + \frac{1}{t _{2005}} =$

$\frac{4004}{3005}$

$\frac{3005}{4004}$

$\frac{8020}{6024}$

$\frac{8008}{6015}$

$\frac{1}{t _{n}} = \frac{4}{( n + 2 ) ( n + 3 )} = \frac{4}{n + 2} - \frac{4}{n + 3}$
Now $(\frac{4}{3} - \frac{4}{4}) + (\frac{4}{4} - \frac{4}{5}) + (\frac{4}{5} - \frac{4}{6}) + ....$
$+ (\frac{4}{2007} - \frac{4}{2008}) =$
$\frac{4}{3} - \frac{4}{2008} = \frac{8020}{6024}$

Q. If tn​=41​(n+2)(n+3) for n = 1,2,3... then t1​1​+t2​1​+t3​1​+....+t2005​1​=

30054004​

40043005​

60248020​

60158008​