Let P(n):4n+15n−1 is divisible by 9 . For n=1,P(1)=41+15×1−1=18, which is divisible by 9 . ∴P(l) is true. Let P(k) be true. Then, 4k+15k−1 is divisible by 9 . ⇒4k+15k−1=9λ, for some λ∈N. We shall now show that P(k+1) is true. For this, we have to show that 4k+1+15(k+1)−1 is divisible by 9 . Now, 4k+1+15(k+1)−1=4k⋅4+15(k+1)−1 =(9λ−15k+1)×4+15(k+1)−1 =36λ−45k+18 =9(4λ−5k+2), which is divisible by 9. ∴P(k+1) is true. Thus, P(k) is true ⇒P(k+1) is true. Hence, by PMI, P(n) is true for all n∈N.