Q.Show that $49^n + 16n - 1$ is divisible by 64 for all positive integers 'n'.

Verify the base case n=1, then show the statement for n=k forces it to hold for n=k+1 — the standard mathematical-induction pattern.

Base case (n=1):

$49^1+16(1)-1 = 49+16-1=64$, which is divisible by 64. So the statement holds for n=1.

Inductive hypothesis: Assume the statement holds for $n=k$, i.e.

$49^k+16k-1 = 64m$ for some integer m, so $49^k = 64m-16k+1$.

Inductive step (show true for n=k+1):

$49^{k+1}+16(k+1)-1 = 49\cdot49^k+16k+16-1 = 49\cdot49^k+16k+15$

Substitute $49^k=64m-16k+1$:

$= 49(64m-16k+1)+16k+15$

$= 49\cdot64m - 49\cdot16k+49+16k+15$

$= 3136m - 784k+16k+64$

$= 3136m-768k+64$

Since $3136=64\times49$ and $768=64\times12$:

$= 64(49m-12k+1)$

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