# Principle of Mathematical Induction?

$\large (\ \underbrace{ 66666\ldots6 }_{n \text{ number of 6's}} \ )^2+ \underbrace{88888\ldots8}_{n \text{ number of 8's}}= \underbrace{44444\ldots4}_{2n \text{ number of 4's}}$

Prove the above equation of positive integer $n$.

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Note by Sandeep Bhardwaj
5 years, 2 months ago

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Observe that $6^2 + 8 = 44$. So, it is true for $n=1$.

Let us suppose that for some $k$,

$(\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + \underbrace{8888\ldots8 }_{k \text{ numbers of 8's}} = \underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$

So, Consider

$(\underbrace{ 6666\ldots6 }_{k+1 \text{ numbers of 6's}})^{2} + \underbrace{8888\ldots8 }_{k+1 \text{ numbers of 8's}}$ $=(6 \times 10^{k} +\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + 8 \times 10^{k} +\underbrace{8888\ldots8 }_{k \text{ numbers of 8's}}$ $=36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}} +(\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + 8 \times 10^{k} + \underbrace{8888\ldots8 }_{k \text{ numbers of 8's}}$ $= 36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}} + 8 \times 10^{k} +\underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$ $= 36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \dfrac{6}{9}(10^{k} - 1) + 8 \times 10^{k} +\underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$ $= 44 \times 10^{2k} + \underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$ $= \underbrace{ 4444\ldots4 }_{2k+2 \text{ numbers of 4's}}$

So, by $\text{Principle of Finite Mathematical Induction}$. The given statement is true.

- 5 years, 2 months ago

$\underbrace{ 6666\ldots6 }_{n \text{ numbers of 6's}} = \dfrac{6}{9}(10^n - 1) = \dfrac{2}{3}(10^n - 1)$ $\underbrace{8888\ldots8 }_{n \text{ numbers of 8's}} = \dfrac{8}{9}(10^n - 1)$ $\underbrace{ 4444\ldots4 }_{2n \text{ numbers of 4's}} = \dfrac{4}{9}(10^{2n} - 1)$

So, $(\underbrace{ 6666\ldots6 }_{n \text{ numbers of 6's}})^2 + \underbrace{8888 \ldots8}_{n \text{number of 8's}} = (\dfrac{2}{3}(10^n - 1))^{2} + \dfrac{8}{9}(10^n - 1)$ $=\dfrac{4}{9}(10^{2n} - 2 \times 10^n +1 + 2 \times 10^n -2)$ $=\dfrac{4}{9}(10^{2n} - 1)$ $=\underbrace{4444 \ldots 4}_{2n \text{ number of 4's}}$

Hence, proved.

- 5 years, 2 months ago

Great! Well done!

- 5 years, 2 months ago

Thank u

- 5 years, 2 months ago

For the sake of the title of this note, can you prove this using mathematical induction?

- 5 years, 2 months ago

Hi @Pi Han Goh , I have even posted the solution using Induction. By the way can u please add me as your friend in Facebook. It's request from me.

- 5 years, 2 months ago