Question

${50}^n-49n-1$ is divisible by ${49}^3$ for all positive integer $n$.

• A. True
• B. False

Answer 1

The correct option is B False

Let us write ${50}^n$ as $(1+49{)}^n$

Using Binomial expansion

$(1+49{)}^n=1{+}^n{C}_1(49\left){+}^n{C}_2\right(49{)}^2{+}^n{C}_3(49{)}^3+....$

$⟹{50}^n=1+49n+\frac{n(n-1)}{2}(49{)}^2+\frac{n(n-1)(n-2)}{6}(49{)}^3+....$

$⟹{50}^n-49n-1=\frac{n(n-1)}{2}(49{)}^2+\frac{n(n-1)(n-2)}{6}(49{)}^3+....$

So the Given expression is not divisible by ${49}^3$