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Let \mathrm{a_{n}=}  number of \mathrm{n} -digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them is 0 .

Let \mathrm{b_{n}=}number of such n  digit integers ending with digit 1 , and \mathrm{c_{n}=}  the numner of such n digit integers ending with digit 0 , then for \mathrm{n \geq 3},

Option: 1

\mathrm{a_{n}=b_{n-1}+c_{n-2}}


Option: 2

\mathrm{a_{n}=b_{n-1}+c_{n-1}}


Option: 3

\mathrm{a_{n}=b_{n-2}+c_{n-1}}


Option: 4

\mathrm{a_{n}=b_{n}+c_{n}}


Answers (1)

best_answer

Note that all such numbers begin with 1 .
If number ends in 1 , then there \mathrm{b_{n}=a_{n-1}} such numbers. If number ends in 0 , then \mathrm{(n-1)} th digit cannot be 0 and thus, there are \mathrm{c_{n}=a_{n-2}} such numbers.

\mathrm{\therefore \quad a_{n}=b_{n}+c_{n}=a_{n-1}+a_{n-2}}

 

Posted by

shivangi.shekhar

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