Show that $49^{n} + 16 n - 1$ is divisible by $64$ for all positive integers n ?

Open in App

Solution

${(49)}^{n} + 16 n - 1$ $= {(1 + 48)}^{n} + 16 n - 1$

$= 1 + 48 n + . . . . . 48^{n} + 16 n - 1$

$= 64 n + n C_{2} {(48)}^{2} + n C_{3} {(48)}^{3} + . . . . . . + {(48)}^{n}$

$= 64 (n + n C_{2} {(6)}^{2} + n C_{3} {(6)}^{3} 48 + . . . . . . + {(6)}^{n} {(8)}^{n - 2}$

$∴ 49 n + 16 n - 1$ is divisible by $64$

Suggest Corrections

Similar questions

n \in N; 49^{n} + 16 n - 1

50^{n} - 49 n - 1

49^{3}

n

9^{n + 1} - 8 n - 9

64

50^{n} - 49 n - 1

49^{3}

n

If P(n) : $49 n + 16 n + λ$ is divisible by 64 for $n ϵ N$ is true. then the least negative integral value of $λ$ is

Join BYJU'S Learning Program