Question

Find the \[{{1025}^{th}}\] term in the sequence 1, 22, 4444, 88888888, ...

Answer 1

Hint: Whenever we come across a situation where there is repeated numbers then we must try to convert them to power of 10 to make the solution easy. Here, taking out common number then we will get in the terms of only ‘1’ like
\[1\left( 1 \right),2\left( 11 \right),4\left( 1111 \right),8\left( 11111111 \right),...\]
 Now, multiplying and dividing with ‘9’ so that we can write ‘9’ as (10-1), and ‘99’ as (100-1) and so on. So, we get general form of the sequence in terms of number of that term.

Complete step by step answer:
Now coming to the problem, let us take the sequence
1, 22, 4444, 88888888, …
Attempting same binary operation to each term will have no change.
Now, rewriting the given sequence as follows
1, 2(11), 4(1111), 8(11111111), …
Now multiplying and dividing each term of the sequence by ‘9’
\[\dfrac{1}{9}(9)\], \[\dfrac{2}{9}(99)\], \[\dfrac{4}{9}(9999)\], \[\dfrac{8}{9}(99999999)\], …
The value in brackets in each term can be modified as
\[\dfrac{1}{9}\left( 10-1 \right)\text{, }\dfrac{2}{9}({{10}^{2}}-1),\text{ }\dfrac{4}{9}({{10}^{4}}-1),\text{ }\dfrac{8}{9}({{10}^{8}}-1),\text{ }...\]
The each term can be modified in terms of number of that term as follows\[\dfrac{{{2}^{0}}}{9}({{10}^{{{2}^{0}}}}-1),\text{ }\dfrac{{{2}^{1}}}{9}({{10}^{{{2}^{1}}}}-1),\text{ }\dfrac{{{2}^{2}}}{9}({{10}^{{{2}^{2}}}}-1),\text{ }\dfrac{{{2}^{3}}}{9}({{10}^{{{2}^{3}}}}-1),\text{ }...\]

The generalisation in each term can be written as follows
\[{{1}^{st}}\] term – power of 2 is 0
\[{{2}^{nd}}\] term – power of 2 is 1
\[{{3}^{rd}}\] term – power of 2 is 2 and so on….
So, for any ‘n’ the \[{{n}^{th}}\] term can be written as
\[{{T}_{n}}\text{ = }\dfrac{{{2}^{n-1}}}{9}({{10}^{{{2}^{n-1}}}}-1)\]
This is called the\[{{n}^{th}}\] term of the sequence
As we need \[{{1025}^{th}}\] term we take n = 1025 and substitute in above equation
Now the value of \[{{1025}^{th}}\] term is
\[{{T}_{1025}}\text{ = }\dfrac{{{2}^{1024}}}{9}({{10}^{{{2}^{1024}}}}-1)\]

Note:
 You need to be careful while converting the sequence of each term in terms of ‘n’. Finding the generalised equation plays an important role in the solution. Sometimes students miss the first term and proceeds with the solution that is converting to generalised equation some may miss the following term
\[\dfrac{1}{9}\left( 10-1 \right)=\dfrac{{{2}^{0}}}{9}({{10}^{{{2}^{0}}}}-1)\]