If t_n= 1 4 (n+2)(n+3) for n=1,2,3 , then 1 t₁ + 1 t₂ + + 1 t₂₀₀₃ is equal to

Step-by-step solution

We have, t_n= 1 4 (n+2)(n+3) 1 t_n = 4 (n+2)(n+3) =4 [ 1 n+2 - 1 n+3 ] Put n=1,2,3, , 2003 1 t₁ =4 ( 1 3 - 1 4 ) aligned & 1 t₂ =4 ( 1 4 - 1 5 ) & aligned 1 t₂₀₀₃ =4 ( 1 2005 - 1 2006 ) On adding these terms, we get 1 t₁ + 1 t₂ + 1 t₃ + + 1 t₂₀₀₃ =4 [ 1 3 - 1 2006 ]=4 [ 2006-3 3 2006 ]= 2 2003 3 1003 = 4006 3009