Numbers-6 (CAT-2000)

If a₁ = 1 and a_n+1 = 2a_n+5, n=1,2,… then a₁₀₀ is equal to

(1)(5*2⁹⁹ - 6)      (2) (5*2⁹⁹ + 6)        (3) (6*2⁹⁹ + 5)        (4)(6*2⁹⁹ - 5)

Solution follows here:

Solution:

The better way is to find some values of a_n and check with the given answer options:

Before proceeding further, let us convert the given answer options in terms of n:

a_n =

(1)(5*2^n-1 – 6)      (2)(5*2^n-1 + 6)      (3)(6*2^n-1 + 5)      (4)(6*2^n-1 – 5)

Given a₁ = 1 this is true only for option (4).

 a1 = 6*2^1-1 – 5 = 6*2⁰ – 5 = 6-5 =1

No other option gives us value 1. Hence answer is option(4).

But for our satisfaction let us try for some more values of a_n:

a_n+1 = 2a_n+5 => a₂ = 2a₁+5 = 2+5 = 7; substitute n = 2 in option (4) => a₂ = 6*2¹-5 = 7

yes, verified correct.

a₃ = 2a₂+5 = 2(7)+5 = 19; substitute n = 3 in option (4) => a₃ = 6*2²-5 = 24-5 = 19

its true, verified correct.

Answer (4)