TS EAMCET 6-Aug-2021 Shift 1 Question Paper

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.