Principle of Mathematical Induction?

( 66666…6⏟n number of 6’s )2+88888…8⏟n number of 8’s=44444…4⏟2n number of 4’s\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 nn.

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#Algebra #MathematicalInduction #Proofs #JEE #IITJEE

Note by Sandeep Bhardwaj
5 years, 10 months ago

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Comments

6666…6⏟n numbers of 6’s=69(10n−1)=23(10n−1) \underbrace{ 6666\ldots6 }_{n \text{ numbers of 6's}} = \dfrac{6}{9}(10^n - 1) = \dfrac{2}{3}(10^n - 1) 8888…8⏟n numbers of 8’s=89(10n−1) \underbrace{8888\ldots8 }_{n \text{ numbers of 8's}} = \dfrac{8}{9}(10^n - 1) 4444…4⏟2n numbers of 4’s=49(102n−1) \underbrace{ 4444\ldots4 }_{2n \text{ numbers of 4's}} = \dfrac{4}{9}(10^{2n} - 1)

So, (6666…6⏟n numbers of 6’s)2+8888…8⏟nnumber of 8’s=(23(10n−1))2+89(10n−1) (\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) =49(102n−2×10n+1+2×10n−2)=\dfrac{4}{9}(10^{2n} - 2 \times 10^n +1 + 2 \times 10^n -2) =49(102n−1)=\dfrac{4}{9}(10^{2n} - 1) =4444…4⏟2n number of 4’s=\underbrace{4444 \ldots 4}_{2n \text{ number of 4's}}

Hence, proved.

Surya Prakash - 5 years, 10 months ago

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Great! Well done!

Sandeep Bhardwaj - 5 years, 10 months ago

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Thank u

Surya Prakash - 5 years, 10 months ago

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@Surya Prakash – For the sake of the title of this note, can you prove this using mathematical induction?

Pi Han Goh - 5 years, 10 months ago

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@Pi Han Goh – 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.

Surya Prakash - 5 years, 10 months ago

Observe that 62+8=44 6^2 + 8 = 44. So, it is true for n=1n=1.

Let us suppose that for some kk,

(6666…6⏟k numbers of 6’s)2+8888…8⏟k numbers of 8’s=4444…4⏟2k numbers of 4’s(\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

(6666…6⏟k+1 numbers of 6’s)2+8888…8⏟k+1 numbers of 8’s(\underbrace{ 6666\ldots6 }_{k+1 \text{ numbers of 6's}})^{2} + \underbrace{8888\ldots8 }_{k+1 \text{ numbers of 8's}} =(6×10k+6666…6⏟k numbers of 6’s)2+8×10k+8888…8⏟k 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}} =36×102k+2×6×10k×6666…6⏟k numbers of 6’s+(6666…6⏟k numbers of 6’s)2+8×10k+8888…8⏟k 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}} =36×102k+2×6×10k×6666…6⏟k numbers of 6’s+8×10k+4444…4⏟2k numbers of 4’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}} =36×102k+2×6×10k×69(10k−1)+8×10k+4444…4⏟2k 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}} =44×102k+4444…4⏟2k numbers of 4’s= 44 \times 10^{2k} + \underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}} =4444…4⏟2k+2 numbers of 4’s= \underbrace{ 4444\ldots4 }_{2k+2 \text{ numbers of 4's}}

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

Surya Prakash - 5 years, 10 months ago
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