Question

49^n + 16n -1 is divisible by 64

Answer 1

Solution:

let
$p(n) = {49}^{n} + 16n - 1$
is divisible by 64.

Check for n=1

$p(1) = {49}^{1} + 16(1) - 1 \\ \\ = 49 + 16 - 1 \\ \\ = 64$
which is divisible by 64.

Now let us assume that it is also divisible by 64 when n= k

$p(k) = {49}^{k} + 16k - 1 = 64 \: r \: \: \: \: r \epsilon \: {Z}^{ + }$
now put n= k+1

$p(k + 1) = {49}^{(k + 1)} + 16(k + 1) - 1 \\ \\ manipulate \: these \: terms \: so \\ \: as \: we \: can \: put \: values \: from \: p(k) \\ \\ p(k + 1) = {49}^{k} \times 49 + 16(k + 1) - 1 \\ \\ \\ = 49( {49}^{k} + 16k - 1 - 16k + 1) + 16(k + 1) - 1 \\ \\ = 49(64 \: r - 16k + 1) + 16k + 16 - 1 \\ \\ = 49 \times 64r - 784k +49 + 16k + 15 \\ \\ = 49 \times 64r - 768k + 64 \\ \\ = 64(49r - 12k + 1) \\ \\ p(k + 1) = 64m \: \: \: m \epsilon \: {Z}^{ + }$
Hence the number is divisible by 64 by some positive integer m.

Prove

Answer 2

Step-by-step explanation:

hope it's helpful to you