Question

Let a sequence be defined by $ {{a}_{1}}=3 $ , $ {{a}_{n}}=3{{a}_{n-1}}+1 $ for all $ n>1 $ . Find the first four terms of the sequence.

Answer 1

Hint: The given expression is neither an A.P. nor a G.P.
Find the value of the 2nd term $ {{a}_{2}} $ by using the given relation as $ {{a}_{2}}=3{{a}_{1}}+1 $ , where $ {{a}_{1}}=3 $ . Then the other terms $ {{a}_{3}} $ , $ {{a}_{4}} $ and $ {{a}_{5}} $ can also be found similarly as $ {{a}_{3}}=3{{a}_{2}}+1 $ , $ {{a}_{4}}=3{{a}_{3}}+1 $ and $ {{a}_{5}}=3{{a}_{4}}+1 $ .
In simpler terms, multiply a term with 3 and add 1 to it to get the next term.

Complete step-by-step answer:
It is given that $ {{a}_{1}}=3 $ .
Using the given (recurrence) relation $ {{a}_{n}}=3{{a}_{n-1}}+1,\ n>1 $ , we can write:
  $ {{a}_{2}}=3{{a}_{1}}+1 $
Substituting the value of $ {{a}_{1}} $ , we get:
⇒ $ {{a}_{2}}=3(3)+1 $
⇒ $ {{a}_{2}}=10 $
Similarly, we can calculate:
  $ {{a}_{3}}=3{{a}_{2}}+1 $
Substituting the value of $ {a_2} $ , we get:
⇒ $ {{a}_{3}}=3(10)+1 $
⇒ $ {{a}_{3}}=31 $
And,
  $ {{a}_{4}}=3{{a}_{3}}+1 $
Substituting the value of $ {a_3} $ , we get:
⇒ $ {{a}_{4}}=3(31)+1 $
⇒ $ {{a}_{4}}=94 $
The value of $ {{a}_{5}} $ can also be calculated as:
  $ {{a}_{5}}=3{{a}_{4}}+1 $
Substituting the value of $ {a_4} $ , we get:
⇒ $ {{a}_{5}}=3(94)+1 $
⇒ $ {{a}_{5}}=283 $
The first few terms of the series are 3, 10, 31, 94, 283, ... etc.
So, the correct answer is “3, 10, 31, 94, 283, ... etc.”.

Note: The representation of the general relation between the terms of a series, which follows a fixed pattern, is called the recurrence relation of the series.
The general form of the recurrence relation of an A.P. is: $ {{a}_{n+1}}={{a}_{n}}+k $ .
The general form of the recurrence relation of a G.P. is: $ {{a}_{n+1}}=k{{a}_{n}} $ .