Explanation

(d)
$49^{n} + 16n - 1 = (1 + 48)^{n} + 16nn - 1$
$= 1 + n_{C_{1}}(48) + n_{C_{2}}(48)^{2} +$….$+ n_{C_{n}}(48)^{n} + 16n - 1$
$= (48n + 16n) + n_{C_{2}}{(48)}^{2} + n_{C_{3}}{(48)}^{3} +$…$+ n_{C_{n}}{(48)}^{n}$
$64n + 8^{2}(n_{C_{2}} \bullet 6^{2} + n_{C_{3}} \bullet 6^{3} \bullet 8 + n_{C_{4}} \bullet 6^{4} \bullet 8^{2} +$……$+ n_{C_{n}} \bullet 6^{n} \bullet 8^{n - 2})$
Hence, $49^{n} + 16n - 1$ is divisible by 64
Alternate Let $P(n) = 49^{n} + 16n - 1$
For $n = 1$
$P(1) = 49 + 16 - 1 = 64$