Question 13

Question 13 14,

Dear Student,

The solution is as under:

$$\begin{gathered} a={\left(6\right)}_n=66666......n \\ b={\left(8\right)}_n=88888......n \\ c={\left(4\right)}_{2n}=44444......2n \\ Let, n=1 \\ so, a=6, b=8, and c=44 \\ Hence, a^2=6^2=36 \\ thus, a^2+b=36+8=44 \\ hence, a^2+b=c \\ or, a^2+b-c=0 \\ Let, n=2 \\ so, a=66, b=88, and c=4444 \\ Hence, a^2={66}^2=4356 \\ thus, a^2+b=4356+88=4444 \\ hence, a^2+b=c \\ or, a^2+b-c=0 \\ Hence, a^2+b={\left(6666.....n\right)}^2+\left(88888......n\right)=\left(44444......2n\right) \\ or, a^2+b=c \\ thus, a^2+b-c=0 \end{gathered}$$

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