Prove by using the principle of mathematical induction that for all `n in N, 10^(n)+(3xx4^(n+2))+5` is divisible by `9`.

Explanation

Let f(n)=`10^(n)+3. 4^(n+2)+5`
Let P(n): f(n) is divisible by 9.
f(1)=`10^(1)+3.4^(1+2)+5`
=10+192+5=207, which is divisible by 9.
`therefore P(1)` is true. (a)
`Rightarrow f(k)=10^(k)+3.4^(k+2)+5=9m, k in N` [where m is any integer]
To prove P(k+1) is true, i.e. f(k+1) is divisible by 9.
Now, f(k+1)=`10^(k+1)+3.4^(k+1)+2=5`
`=10^(k).10+3.4^(k+3)+5`
`=90m-18.4^(k+2)-5)` which is divisible by 9.
Hence P(k+1) is true whenever P(k) is true. .....(b)
From (a) and (b), by the principle of mathematical induction it follows that P(n) is true for all `n in NN`.