Given, $$b_n$$ denotes the number of $$n$$-digit integer formed by the digits 0, 1 or both such that $$n$$-digit integer ending with 1 and no consecutive digits are '0'.

$$\therefore \quad b_6=$$ six digit number ending with 1.

Like 1 ........... 1, and rest four places are filled as Case No. (I) : Use four ' 1 '

$$\text { Case No. (I) : Use four ' } 1 \text { ' }$$

$$\underline 1 \underbrace {\underline 1 \,\underline 1 \,\underline 1 \,\underline 1 }_{}\underline 1 $$

Number of ways = 1

Case No. (II) : Use three '1' and one '0'

$$\underline 1 \underbrace {\underline 1 \,\underline 1 \,\underline 1 \,\underline 0 }_{}\underline 1 $$

Number of ways $$=\frac{4!}{3!}=4$$

Case No. (III) : Use two '1' and two '0'

$$\underline 1 \underbrace {\underline 1 \,\underline 0 \,\underline 1 \,\underline 0 }_{}\underline 1 $$

or

$$\underline 1 \underbrace {\underline 1 \,\underline 1 \,\underline 0 \,\underline 1 }_{}\underline 1 $$

or

$$\underline 1 \underbrace {\underline 0 \,\underline 1 \,\underline 1 \,\underline 0 }_{}\underline 1 $$

No. of ways = 3

Hence, $$b_6=1+4+3=8$$