Mathematics - Binomial Theorem Question with Solution | TestHub

Given,

$\left(S1\right) : (2023{)}^{2022}-(1999{)}^{2022}$ is divisible by $8$

Now we know that $(x-y)$ divides $\left(x^n-y^n\right) \forall n\in \mathrm{\mathbb{N}}$

So, $\left(2023-1999\right)$ divides $(2023{)}^{2022}-(1999{)}^{2022}$

$\Rightarrow 24$ divides $(2023{)}^{2022}-(1999{)}^{2002}$

$\Rightarrow 8$ will divide $(2023{)}^{2022}-(1999{)}^{2002}$

As $8$ divides $24$

Hence, $\left(S1\right)$ is correct

Now solving,

$\left(S2\right) : 13{\left(13\right)}^n-11n-13$ is divisible by $144$ for $n\in \mathrm{\mathbb{N}}$

So using binomial theorem in ${\left(1+12\right)}^n$ we get,

$13{\left(1+12\right)}^n-11n-13$

$=13\left({}^n{C}_0{+}^n{C}_{1}12{+}^n{C}_2{12}^2+...{+}^n{C}_n{12}^n\right)-11n-13$

$=12\times 13n-11n+{12}^2\lambda$

$=145n+144\lambda$ which is not divisible by $144$

Hence, $\left(S2\right)$ is incorrect