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Grade: 12th pass
        
what is the1025th term in series 1,22,4444,88888888?
7 months ago

Answers : (1)

Arun
12335 Points
							

HI,

Its a very interesting series and complicated too..

If we observe the 1st term that is 1, it has 2^0 no of digits that is 1 digit.

If we observe the 2nd term that is 22, it has 2^1 no of digits that is 2 digits.

similarly if we observe the 3rd term that is 4444, it has 2^2 no of digits that is 4 digits.

And the 4th term contains 2^3 no of digits , that is 8 digits.

Hence each term contains 2^(n-1) no of digits and each digit is equal to the number of digits.

This is important:- EACH DIGIT IS EQUAL TO THE NUMBER OF DIGITS.

The nth term then would have 2^(n-1) digits with each digit equal to 2^(n-1).

Hence the 1025th term would contain 2^(1025–1) terms, that is 2^1024 digits, each digit being equal to 2^1024.

So, the 1025th term would be something like this:-

(2^1024)2^(1024)2^(1024)…………………1024 times.

Now lets proceed to check the value of this huge number by a simple algebraic method. Lets consider a number whose all digits are same.

Suppose a number is 999. We can write this number as:-

999=9(100+10+1).

999=9(10^2+10^1+10^0) that is in a from of GP.

Similarly our 1025th term , we can write it as :-

2^(1024)2^(1024)2^(1024)…….1024 times=2^1024(10^1023+10^1022+…10^0).

Sum of a GP is (r^n-1)/(r-1).

where r is the common ratio and n is the number of terms.

Hence our 1025th term =2^1024[(10^1024–1)/(10–1)]=2^1024[(10^1024–1)/9].

7 months ago
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