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Question Number 67501 by TawaTawa last updated on 28/Aug/19

Show that 1n^3 + 2n + 3n^2 is divisible by 2 and 3 for all positive integers n.

$$\mathrm{Show}\:\mathrm{that}\:\:\mathrm{1n}^{\mathrm{3}} \:+\:\mathrm{2n}\:+\:\mathrm{3n}^{\mathrm{2}} \:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{n}. \\ $$

Commented by Prithwish sen last updated on 28/Aug/19

Another approch We know that (1+x)^n =1+nx+ ((n(n−1))/(2!))x^2 + ((n(n−1)(n−2))/(3!))x^(3 +) ........+((n(n−1)(n−2).......(n−r+1))/(r!)) x^r +.....+x^n Now as far n∈N the coefficients of the expression are integers ∴ n(n−1) must be divisible by 2! = 2 n(n−1)(n−2) ′′ ′′ ′′ ′′ 3! = 6 ............................... n(n−1)(n−2)....(n−r+1) must be divisible by r! Now by putting n = n+2 at n(n−1)(n−2) we get , n(n+1)(n+2) must be divisible by 3! = 6 proved.

$$\mathrm{Another}\:\mathrm{approch} \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that} \\ $$$$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} =\mathrm{1}+\mathrm{nx}+\:\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}!}\mathrm{x}^{\mathrm{2}} \:+\:\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{3}!}\mathrm{x}^{\mathrm{3}\:+} \\ $$$$........+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right).......\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)}{\mathrm{r}!}\:\mathrm{x}^{\mathrm{r}} +.....+\mathrm{x}^{\mathrm{n}} \\ $$$$\mathrm{Now}\:\mathrm{as}\:\mathrm{far}\:\mathrm{n}\in\mathbb{N}\:\mathrm{the}\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{are}\:\mathrm{integers} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\:\:\:\:\:\boldsymbol{\mathrm{must}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{divisible}}\:\boldsymbol{\mathrm{by}}\:\mathrm{2}!\:=\:\mathrm{2} \\ $$$$\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}−\mathrm{2}\right)\:\:\:\:\:\:\:\:''\:\:\:\:\:\:\:\:\:''\:\:\:\:\:\:\:\:''\:\:\:\:\:\:\:\:\:\:\:\:\:''\:\:\:\mathrm{3}!\:=\:\mathrm{6} \\ $$$$............................... \\ $$$$\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}−\mathrm{2}\right)....\left(\boldsymbol{\mathrm{n}}−\boldsymbol{\mathrm{r}}+\mathrm{1}\right)\:\:\boldsymbol{\mathrm{must}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{divisible}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{r}}! \\ $$$$\mathrm{Now}\:\mathrm{by}\:\mathrm{putting}\:\boldsymbol{\mathrm{n}}\:=\:\boldsymbol{\mathrm{n}}+\mathrm{2}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}−\mathrm{2}\right)\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\:, \\ $$$$\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)\:\boldsymbol{\mathrm{must}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{divisible}}\:\boldsymbol{\mathrm{by}}\:\mathrm{3}!\:=\:\mathrm{6}\:\boldsymbol{\mathrm{proved}}. \\ $$

Commented by TawaTawa last updated on 28/Aug/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 28/Aug/19

n^3 +3n^2 +2n=n(n+1)(n+3) For 2:n∈E ∨ n∈O n ∈E⇒ 2 ∣ n⇒2 ∣ n(n+1)(n+2) ⇒2 ∣ n^3 +3n^2 +2n n∈O⇒n+1∈E⇒2 ∣ n+1 ⇒2 ∣ n(n+1)(n+2) ⇒2 ∣ n^3 +3n^2 +2n For 3:n=3k or n=3k+1 or n=3k+2 ∀ k∈Z n=3k⇒3∣n⇒3∣n(n+1)(n+2) ⇒3∣n^3 +3n^2 +2n n=3k+1⇒n+2=3k+1+2=3(k+1) 3∣n+2⇒3∣n(n+1)(n+2) ⇒3∣n^3 +3n^2 +2n n=3k+2⇒n+1=3k+3=3(k+1) ⇒3∣n+1⇒3∣n(n+1)(n+2) ⇒3∣n^3 +3n^2 +2n Hence 2∣n^3 +3n^2 +2n ∧ 3∣n^3 +3n^2 +2n

$${n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right) \\ $$$${For}\:\mathrm{2}:{n}\in\mathbb{E}\:\vee\:{n}\in\mathbb{O} \\ $$$$\:{n}\:\in\mathbb{E}\Rightarrow\:\:\mathrm{2}\:\mid\:{n}\Rightarrow\mathrm{2}\:\mid\:{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{2}\:\mid\:{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n} \\ $$$${n}\in\mathbb{O}\Rightarrow{n}+\mathrm{1}\in\mathbb{E}\Rightarrow\mathrm{2}\:\mid\:{n}+\mathrm{1}\:\Rightarrow\mathrm{2}\:\mid\:{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{2}\:\mid\:{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n} \\ $$$${For}\:\mathrm{3}:{n}=\mathrm{3}{k}\:{or}\:{n}=\mathrm{3}{k}+\mathrm{1}\:{or}\:{n}=\mathrm{3}{k}+\mathrm{2}\:\:\:\:\:\forall\:{k}\in\mathbb{Z} \\ $$$${n}=\mathrm{3}{k}\Rightarrow\mathrm{3}\mid{n}\Rightarrow\mathrm{3}\mid{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}\mid{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n} \\ $$$${n}=\mathrm{3}{k}+\mathrm{1}\Rightarrow{n}+\mathrm{2}=\mathrm{3}{k}+\mathrm{1}+\mathrm{2}=\mathrm{3}\left(\mathrm{k}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\mid{n}+\mathrm{2}\Rightarrow\mathrm{3}\mid{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}\mid{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n} \\ $$$${n}=\mathrm{3}{k}+\mathrm{2}\Rightarrow{n}+\mathrm{1}=\mathrm{3}{k}+\mathrm{3}=\mathrm{3}\left({k}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}\mid{n}+\mathrm{1}\Rightarrow\mathrm{3}\mid{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}\mid{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n} \\ $$$${Hence}\:\mathrm{2}\mid{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}\:\wedge\:\mathrm{3}\mid{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n} \\ $$$$ \\ $$

Commented by TawaTawa last updated on 28/Aug/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by petrochengula last updated on 28/Aug/19

n^3 +2n+3n^2 =n(n+1)(n+2) let the statement p(n) given as p(n):n^3 +2n+3n^2 is divisible by 2 and 3, ∀nεz^+ we observe that p(1) is true, since 1^2 +2+3=6 is divisible by 2 and 3 assume that p(n) is true for some integer k p(k):k^3 +2k+3k^(2 ) is divisible by 2 and 3 such that k^3 +2k+3k^2 =2p and k^3 +2k+3k^2 =3q where p,qεZ^+ . Now to prove that p(k+1):(k+1)^3 +2(k+1)+3(k+1)^2 1^(st ) case we have to show that p(k+1) is divisible by 2 (k+1)^3 +2(k+1)+3(k+1)^2 =k^3 +3k^2 +3k+2k+2+3(k^2 +2k+1) =k^3 +2k+3k^2 +2(k^2 +3k+2)+(k+1)(k+2) but k^3 +2k+3k^2 =2p ⇒k(k+1)(k+2)=2p⇒(k+1)(k+2)=((2p)/k) (k+1)^3 +2(k+1)+3(k+1)^2 =2p+2(k^2 +3k+2)+((2p)/k) =2(p+k^2 +3k+2+(p/k))=2m 2^(nd ) case We have to show that p(k+1) is divisible by 3 (k+1)^3 +2(k+1)+3(k+1)^2 =k^3 +2k+3k^2 +3k^2 +9k+6 =3q+3(k^2 +3k+2) =3(q+k^2 +3k+2) Thus p(k+1) is true,whenever p(k) is true Hence by the principle of mathematical induction p(n) is true for all positive integers number n.

$${n}^{\mathrm{3}} +\mathrm{2}{n}+\mathrm{3}{n}^{\mathrm{2}} ={n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$${let}\:{the}\:{statement}\:{p}\left({n}\right)\:{given}\:{as}\:{p}\left({n}\right):{n}^{\mathrm{3}} +\mathrm{2}{n}+\mathrm{3}{n}^{\mathrm{2}} \:{is}\:{divisible}\:{by}\:\mathrm{2}\:{and}\:\mathrm{3},\:\forall{n}\epsilon{z}^{+} \\ $$$${we}\:{observe}\:{that}\:{p}\left(\mathrm{1}\right)\:{is}\:{true},\:{since}\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}+\mathrm{3}=\mathrm{6}\:{is}\:{divisible}\:{by}\:\mathrm{2}\:{and}\:\mathrm{3} \\ $$$${assume}\:{that}\:{p}\left({n}\right)\:{is}\:{true}\:{for}\:{some}\:{integer}\:{k} \\ $$$${p}\left({k}\right):{k}^{\mathrm{3}} +\mathrm{2}{k}+\mathrm{3}{k}^{\mathrm{2}\:} \:{is}\:{divisible}\:{by}\:\mathrm{2}\:{and}\:\mathrm{3}\:{such}\:{that}\:{k}^{\mathrm{3}} +\mathrm{2}{k}+\mathrm{3}{k}^{\mathrm{2}} =\mathrm{2}{p}\:{and}\:{k}^{\mathrm{3}} +\mathrm{2}{k}+\mathrm{3}{k}^{\mathrm{2}} =\mathrm{3}{q}\:{where}\:{p},{q}\epsilon{Z}^{+} .\:{Now}\:{to}\:{prove}\:{that}\:{p}\left({k}+\mathrm{1}\right):\left({k}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}^{{st}\:} {case} \\ $$$${we}\:{have}\:{to}\:{show}\:{that}\:{p}\left({k}+\mathrm{1}\right)\:{is}\:{divisible}\:{by}\:\mathrm{2} \\ $$$$\left({k}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{2}} ={k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}{k}+\mathrm{2}+\mathrm{3}\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$={k}^{\mathrm{3}} +\mathrm{2}{k}+\mathrm{3}{k}^{\mathrm{2}} +\mathrm{2}\left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}\right)+\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right) \\ $$$${but}\:{k}^{\mathrm{3}} +\mathrm{2}{k}+\mathrm{3}{k}^{\mathrm{2}} =\mathrm{2}{p}\:\Rightarrow{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)=\mathrm{2}{p}\Rightarrow\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)=\frac{\mathrm{2}{p}}{{k}} \\ $$$$\left({k}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}{p}+\mathrm{2}\left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}\right)+\frac{\mathrm{2}{p}}{{k}} \\ $$$$=\mathrm{2}\left({p}+{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}+\frac{{p}}{{k}}\right)=\mathrm{2}{m} \\ $$$$\mathrm{2}^{{nd}\:} {case} \\ $$$${We}\:{have}\:{to}\:{show}\:{that}\:{p}\left({k}+\mathrm{1}\right)\:{is}\:{divisible}\:{by}\:\mathrm{3} \\ $$$$\left({k}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{3}\left({k}+\mathrm{1}\right)^{\mathrm{2}} ={k}^{\mathrm{3}} +\mathrm{2}{k}+\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}^{\mathrm{2}} +\mathrm{9}{k}+\mathrm{6} \\ $$$$=\mathrm{3}{q}+\mathrm{3}\left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}\right) \\ $$$$=\mathrm{3}\left({q}+{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}\right) \\ $$$${Thus}\:{p}\left({k}+\mathrm{1}\right)\:{is}\:{true},{whenever}\:{p}\left({k}\right)\:{is}\:{true} \\ $$$${Hence}\:{by}\:{the}\:{principle}\:{of}\:{mathematical}\:{induction}\:{p}\left({n}\right)\:{is}\:{true}\:{for}\:{all}\:{positive}\:{integers}\:{number}\:{n}. \\ $$

Commented by TawaTawa last updated on 28/Aug/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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