Question
By the principle of induction $$\forall n \in N,{3^{2n}}$$ when divided by 8, leaves remainder
A.
2
B.
3
C.
7
D.
1
Answer :
1
Solution :
Let $$P\left( n \right)$$ be the statement given by
$$P\left( n \right):{3^{2n}}$$ when divided by 8, the remainder is 1.
or $$P\left( n \right):{3^{2n}} = 8\lambda + 1$$ for some $$\lambda \in N$$
For $$n = 1,P\left( 1 \right):{3^2} = \left( {8 \times 1} \right) + 1 = 8\lambda + 1,\,$$ where $$\lambda = 1$$
$$\eqalign{
& \therefore P\left( 1 \right){\text{ is true}}{\text{.}} \cr
& {\text{Let, }}P\left( k \right){\text{ be true}}{\text{.}} \cr} $$
Then, $${{\text{3}}^{2k}} = 8\lambda + 1$$ for some $$\lambda \in N\,\,\,\,.....\left( {\text{i}} \right)$$
We shall now show that $$P\left( {k + 1} \right)$$ is true, for which we have to show that $${{\text{3}}^{2\left( {k + 1} \right)}}$$ when divided by 8, the remainder is 1.
$$\eqalign{
& {\text{Now, }}{{\text{3}}^{2\left( {k + 1} \right)}} = {{\text{3}}^{2k}} \cdot {{\text{3}}^2} \cr
& = \left( {8\lambda + 1} \right) \times 9\,\,\,\left[ {{\text{Using }}\left( {\text{i}} \right)} \right] \cr
& = 72\lambda + 9 = 72\lambda + 8 + 1 = 8\left( {9\lambda + 1} \right) + 1 \cr
& = 8\mu + 1,{\text{ where }}\mu = 9\lambda + 1 \in N \cr
& \Rightarrow P\left( {k + 1} \right){\text{ is true}}{\text{.}} \cr} $$
Thus, $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( {k} \right)$$ is true.
Hence, by the principle of mathematical induction $$P\left( {n} \right)$$ is true for all $$n \in N.$$