Question

${49}^n+16n-1$ is divisible by

• A. $3$
• B. $29$
• C. $19$
• D. $64$

Answer 1

Answer: (D)

$64$

Now, ${49}^n+16n-1=(1+48{)}^n+16n-1$
$=1{+}^n{C}_1\left(48\right){+}^n{C}_2(48{)}^2+........{+}^n{C}_n(48{)}^n+16n-1$
$=(48n+16n){+}^n{C}_2\left(48{)}^2{+}^n{C}_3\right(48{)}^3+..{+}^n{C}_n(48{)}^n$
$=64n+8^2{(}^n{C}_2{6}^2{+}^n{C}_3.6^3.8{+}^n{C}_4.6^4.8^2+{+}^n{C}_n{.6}^n{8}^{n-2})$
Hence, ${49}^n+16n-1$ is divisible by $64$.