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`        what is the1025th term in series 1,22,4444,88888888?`
one year ago

Arun
23006 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].
```
one year ago
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