Question

$\sqrt{10}\{{(\sqrt{10}+1)}^{100}-{(\sqrt{10}-1)}^{100}\}$ is whole number.

• A. True
• B. False

Answer 1

The correct option is A True

Given number is,

$\sqrt{10}[{(\sqrt{10}+1)}^{100}-{(\sqrt{10}-1)}^{100}]$

$=\sqrt{10}[{\left(\sqrt{10}(1+\frac{1}{\sqrt{10}})\right)}^{100}-{\left(\sqrt{10}(1-\frac{1}{\sqrt{10}})\right)}^{100}]$

$=\sqrt{10}[{10}^{50}{(1+\frac{1}{\sqrt{10}})}^{100}-{10}^{50}{(1-\frac{1}{\sqrt{10}})}^{100}]$

$=\sqrt{10}\times {10}^{50}[{(1+\frac{1}{\sqrt{10}})}^{100}-{(1-\frac{1}{\sqrt{10}})}^{100}]$

$=\sqrt{10}\times {10}^{50}[(1+100.\frac{1}{\sqrt{10}}+\frac{100\times 99}{2!}{\left(\frac{1}{\sqrt{10}}\right)}^2+\frac{100\times 99\times 98}{3!}{\left(\frac{1}{\sqrt{10}}\right)}^3+......)-(1-100.\frac{1}{\sqrt{10}}+\frac{100\times 99}{2!}{\left(\frac{1}{\sqrt{10}}\right)}^2-\frac{100\times 99\times 98}{3!}{\left(\frac{1}{\sqrt{10}}\right)}^3+......)]$

$=\sqrt{10}\times {10}^{50}[(1+100.\frac{1}{\sqrt{10}}+\frac{100\times 99}{2!}{\left(\frac{1}{\sqrt{10}}\right)}^2+\frac{100\times 99\times 98}{3!}{\left(\frac{1}{\sqrt{10}}\right)}^3+......)-1+100.\frac{1}{\sqrt{10}}-\frac{100\times 99}{2!}{\left(\frac{1}{\sqrt{10}}\right)}^2+\frac{100\times 99\times 98}{3!}{\left(\frac{1}{\sqrt{10}}\right)}^3-......]$

It is clear that all the odd powers get cancelled with each other.

$=\sqrt{10}\times {10}^{50}[100.\frac{1}{\sqrt{10}}+\frac{100\times 99\times 98}{3!}{\left(\frac{1}{\sqrt{10}}\right)}^3+......+100.\frac{1}{\sqrt{10}}+\frac{100\times 99\times 98}{3!}{\left(\frac{1}{\sqrt{10}}\right)}^3-......]$

$=\sqrt{10}\times {10}^{50}[\frac{200}{\sqrt{10}}+\frac{2\times 100\times 99\times 98}{3!}\frac{1}{10\sqrt{10}}]$

$={10}^{50}[200+(10\times 33\times 98)+.......]$

All the numbers in this expression are whole numbers.

Thus, given number is whole number.