Question

Prove that by using principle of mathematical induction 4^n+15n+1 is divisible by 9 for all natural no. N

Answer 1

AnswEr:

Let P(n) be the statement given by

$\qquad\tt{P(n)\colon\:4^n+15n-1\:is\:divisible\:by\:9}$

Step I

$\tt \: p(1) = {4}^{1} + 15 \times 1 - 1 \: \: \sf is \: divisible \: by \: 9 \\ \\ \tt \degree. \degree {4}^{1} + 15 \times 1 - 1 = 18 \: \: \sf \: which \: is \: divisible \\ \sf \: by \: 9 \\ \\ \therefore \sf p(1) \: is \: true$

Step II

Let P(m) be true. Then,

$\qquad\tt{4^m+15m-1\:\:is\:divisible\:by\:9}$

$\implies \sf {4}^{m} + 15m - 1 = 9 \lambda \: \: \: for \: some \\ \sf \lambda \: \in \: N \:$

________________________

$\star \: \rm We \: shall \: now \: show \: that \\ \rm P(m + 1) \: is \: true, \: for \: this \: we \\ \rm \: have \: to \: show \: that \: {4}^{m + 1} + 15(m +1 ) - 1 \\ \rm \: is \: divisible \: by \: 9$

Now,

$\tt {4}^{m + 1} + 15(m + 1) = 1 = {4}^{m} .4 + 15(m + 1) - 1 \\ \\ \tt =( 9 \lambda - 15m + 1) \times 4 + 15(m + 1) - 1 \\ \\ \tt \: = 36 \lambda - 45m + 18 = 9(4 \lambda - 5m + 2) \\ , \rm \: which \: is \: divisible \: by \: 9. \\ \\ \therefore \sf \: P(m + 1) \: is \: true.$

Thus, P(m) is true => P(m+1) is true.

Hence, by the principle of mathematical induction P(n) is true for all n $\in$ N i.e., $\sf{4^n+15n-1}$ is divisible by 9.