# By-Induction-Prove-that-12-n-4-n-2-

Question Number 2857 by Syaka last updated on 29/Nov/15
$${By}\:{Induction}\:{Prove}\:{that}\:: \\$$$$\\$$$$\mathrm{12}\mid\left({n}^{\mathrm{4}} \:−\:{n}^{\mathrm{2}} \right) \\$$
Commented by Filup last updated on 29/Nov/15
$$\mathrm{What}\:\mathrm{does}\:\mid\:\mathrm{mean}? \\$$
Commented by prakash jain last updated on 29/Nov/15
$${n}^{\mathrm{4}} −{n}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{12}. \\$$
Commented by Syaka last updated on 29/Nov/15
$${yes},\:{Sir} \\$$
Answered by prakash jain last updated on 29/Nov/15
$${f}\left({n}\right)={n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)=\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right){n} \\$$$$\mathrm{Statement}\:\mathrm{1}. \\$$$${g}\left({n}\right)=\left({n}−\mathrm{1}\right){n}\:\mathrm{is}\:\mathrm{divisible}\:{by}\:\mathrm{2}.\:\:\:…\left(\mathrm{A}\right) \\$$$${g}\left(\mathrm{1}\right)=\mathrm{0}, \\$$$${g}\left(\mathrm{2}\right)=\mathrm{2} \\$$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{2}\mid{g}\left({n}\right)\:\mathrm{for}\:\mathrm{some}\:{n}\:\mathrm{so}\:{n}\left({n}−\mathrm{1}\right)=\mathrm{2}{k} \\$$$${g}\left({n}+\mathrm{1}\right)=\left({n}+\mathrm{1}\right){n}={n}^{\mathrm{2}} +{n}={n}^{\mathrm{2}} −{n}+\mathrm{2}{n}=\mathrm{2}\left({k}+{n}\right) \\$$$$\mathrm{So}\:\mathrm{the}\:\mathrm{statement}\:\left(\mathrm{A}\right)\:\mathrm{is}\:\mathrm{true}\:\forall{n}. \\$$$$\mathrm{It}\:\mathrm{follows}\:\mathrm{that}\:{h}\left({n}\right)={n}\left({n}+\mathrm{1}\right)\: \\$$$$\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}.\:\:\:\: \\$$$${f}\left({n}\right)={g}\left({n}\right){h}\left({n}\right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{4}.\:\:\:\:\:\:…\left({B}\right) \\$$$$\mathrm{Statement}\:\mathrm{2}: \\$$$${y}\left({n}\right)=\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divisible}\:{by}\:\mathrm{3}.\:\:…\left(\mathrm{C}\right) \\$$$${y}\left(\mathrm{1}\right)=\mathrm{0} \\$$$$\mathrm{Let}\:\mathrm{us}\:{y}\left({n}\right)\:\mathrm{is}\:\mathrm{divisible}\:{by}\:\mathrm{3}\:\mathrm{for}\:\mathrm{some}\:{n}. \\$$$${y}\left({n}\right)=\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)=\mathrm{3}{j} \\$$$${y}\left({n}+\mathrm{1}\right)={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)={n}\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{3}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={n}\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)=\mathrm{3}{j}+\mathrm{3}{n}\left({n}+\mathrm{1}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\left(\mathrm{j}+{n}^{\mathrm{2}} +{n}\right) \\$$$${H}\mathrm{ence}\:{y}\left({n}\right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}\:\mathrm{for}\:\mathrm{all}\:{n}. \\$$$$\mathrm{From}\:\mathrm{statement}\:\left(\mathrm{B}\right)\:\mathrm{and}\:\left(\mathrm{C}\right) \\$$$${f}\left({n}\right)=\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right){n}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{4}×\mathrm{3}=\mathrm{12} \\$$$$\mathrm{since}\:\mathrm{3}\:\mathrm{and}\:\mathrm{4}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{factor}. \\$$
Commented by RasheedAhmad last updated on 29/Nov/15
$$\mathrm{6}{th}\:{line}:{n}\left({n}−\mathrm{1}\right)=\mathrm{2}{k} \\$$$$\mathrm{7}{th}\:{line}:{n}^{\mathrm{2}} −{n}+\mathrm{2}{n}=\mathrm{2}\left({k}+\mathrm{1}\right)? \\$$$${n}^{\mathrm{2}} −{n}+\mathrm{2}{n}=\mathrm{2}{k}+\mathrm{2}{n}=\mathrm{2}\left({k}+{n}\right) \\$$
Commented by prakash jain last updated on 29/Nov/15
$$\mathrm{Thanks}\:\mathrm{Rasheed}. \\$$$$\mathrm{I}\:\mathrm{have}\:\mathrm{corrected}\:\mathrm{the}\:\mathrm{mistake}. \\$$
Commented by Rasheed Soomro last updated on 29/Nov/15
$${g}\left({n}+\mathrm{1}\right)=\left({n}+\mathrm{1}\right){n}={n}^{\mathrm{2}} +{n}={n}^{\mathrm{2}} −{n}+\mathrm{2}{n}=\mathrm{2}\left({k}+{n}\right) \\$$$$\mathrm{So}\:\mathrm{the}\:\mathrm{statement}\:\left(\mathrm{A}\right)\:\mathrm{is}\:\mathrm{true}\:\forall{n}. \\$$$${How}\:{is}\:\mathrm{2}\left({k}+{n}\right)\:{of}\:{the}\:{type}\:\left({n}−\mathrm{1}\right){n} \\$$$${g}\left({n}\right)=\left({n}−\mathrm{1}\right){n} \\$$$${g}\left({n}+\mathrm{1}\right)=\mathrm{2}\left({k}+{n}\right) \\$$$$\mathcal{O}{n}\:{lhs}\:{n}+\mathrm{1}\:{indicates}\:{that}\:{you}\:{are}\:{seeking}\: \\$$$${result}\:{for}\:{n}+\mathrm{1},{but}\:{on}\:{rhs}\:{you}\:{reached}\:{at}\:\mathrm{2}\left({k}+{n}\right) \\$$$$\left[{instead}\:{of}\:\left(\overline {{n}+\mathrm{1}}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)\right] \\$$$${Why}\:\mathrm{So}\:\mathrm{the}\:\mathrm{statement}\:\left(\mathrm{A}\right)\:\mathrm{is}\:\mathrm{true}\:\forall{n}.? \\$$
Commented by prakash jain last updated on 29/Nov/15
$${g}\left({n}\right)={n}\left({n}−\mathrm{1}\right)={n}^{\mathrm{2}} −{n}\: \\$$$${g}\left(\mathrm{1}\right)=\mathrm{0}\:{so}\:{g}\left(\mathrm{1}\right)\:{is}\:{divisible}\:{by}\:\mathrm{2}. \\$$$${assume}\:{if}\:{g}\left({n}\right)\:{is}\:{divisible}\:{by}\:\mathrm{2}.\:{g}\left({n}\right)=\mathrm{2}{k} \\$$$${g}\left({n}+\mathrm{1}\right)=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}−\mathrm{1}\right)=\left({n}+\mathrm{1}\right){n} \\$$$$={n}^{\mathrm{2}} +{n}={n}^{\mathrm{2}} −{n}+\mathrm{2}{n}={g}\left({n}−\mathrm{1}\right)+\mathrm{2}{n}=\mathrm{2}{k}+\mathrm{2}{n} \\$$$$=\mathrm{2}\left({k}+{n}\right)\Rightarrow\mathrm{2}\mid{g}\left({n}+\mathrm{1}\right) \\$$$${since}\:\mathrm{2}\mid{g}\left(\mathrm{1}\right)\:{is}\:{true}. \\$$$${if}\:\mathrm{2}\mid{g}\left({n}\right)\:\Rightarrow\mathrm{2}\mid{g}\left({n}+\mathrm{1}\right) \\$$$${hence}\:{g}\left({n}\right)\:{is}\:{divisible}\:{by}\:\mathrm{2}\:{for}\:{all}\:{n}\in\mathbb{N}. \\$$
Commented by Rasheed Soomro last updated on 30/Nov/15
$$\mathcal{T}\boldsymbol{{h}}^{\boldsymbol{{a}}} \boldsymbol{\mathcal{N}{k}}\mathcal{S}\:\mathcal{S}\boldsymbol{{ir}}!\:\mathcal{I}\:\boldsymbol{{underst}}{oo}\boldsymbol{{d}}\:\boldsymbol{{now}}. \\$$
Commented by Rasheed Soomro last updated on 30/Nov/15
$$\boldsymbol{\mathcal{EXCELLENT}}\:\boldsymbol{\mathcal{A}{pproach}}! \\$$