00:02
Hi, we have t of n is equal to 1 if n is equal to 1, 2 into t of n by 3 plus n if n is greater than 2.
00:15
Now let us consider t of n is equal to 2 into t of n by 3 plus n.
00:22
Now replacing n by n by 3 we have t of n by 3 is equal to 2 into p of n by 9 plus n by 3.
00:33
Replacing n by n by 9 we have t of n by 9 is equal to 2 into t of n by 27 plus n by 9 this goes on and now let us consider p of n is equal to 2 into t of n by 3 plus n this is equal to 2 into t of n by 3 plus n this is equal to 2 into now substituting t of n by 3 value we have 2 into t of n by 9 plus n by 3 plus n.
01:12
This is equal to n plus 2 n by 3 plus 2 square into t of n by 9.
01:21
Now replacing t of n by 9 value we have n plus 2 n by 3 plus 2 square n by 9 plus 2 square n by 9 plus 2 cube t of n by 27.
01:38
This can be written as n into 2 by 3 rise to the par 0 plus 2 by 3 rise to the power 1 plus 2 by 3 rise to the part 2 plus 2 rise to the par 3 into t of n by 27.
02:01
Therefore in general we can write p of n is equal to n into 2 by 3 rise to the power 0 plus 2 by 3 rise to the par 1 plus 2 by 3 rise to the par 2 so on up to 2 2 by 3 rise to the par k minus 1 plus 2 rise to the par k into p of n by 3 rise to the par k minus 1 plus 2 rise to the par k into p of n by 3 rise to the par.
02:37
We have another condition given t of 1 is equal to 1.
02:42
This implies n by 3 rise to the par k is equal to 1.
02:49
Implice n is equal to 3 rise to the power k.
02:54
Taking log on both side we have log n is equal to k into log of 3.
03:02
This implies k is equal to log of n with logarithmic base 3...