Explanation

(c)
We have, $49^{n} + 16n - 1 = (1 + 48)^{n} + 16n - 1$
$= 1 + \ ^{n}C_{1}(48) + \ ^{n}C_{2}{(48)}^{2} + ... + \ ^{n}C_{n}{(48)}^{n} + 16n - 1$
$= (48n + 16n) + \ ^{n}C_{2}{(48)}^{2} + \ ^{n}C_{3}{(48)}^{3} + ... + \ ^{n}C_{n}{(48)}^{n}$
$= 64n + 8^{2}\left\lbrack \ ^{n}C_{2} \bullet 6^{2} + \ ^{n}C_{3} \bullet 6^{3} \bullet 8 + \ ^{n}C_{4} \bullet 6^{4} \bullet 8^{2} + ... + \ ^{n}C_{n} \bullet 6^{n} \bullet 8^{n - 2} \right\rbrack$
Hence, $49^{n} + 16n - 1$ is divisible by 64