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Question Number 165329 by mnjuly1970 last updated on 30/Jan/22
prove ( n∈ N ) 3(n+1) ∣ n^( 3) + (n+1)^( 3) + (n+2 )^( 3)
$$ \\ $$$$\:\:\:\:{prove}\:\:\:\:\:\:\:\:\:\:\left(\:{n}\in\:\mathbb{N}\:\right) \\ $$$$\:\:\:\:\mathrm{3}\left({n}+\mathrm{1}\right)\:\mid\:{n}^{\:\mathrm{3}} \:+\:\left({n}+\mathrm{1}\right)^{\:\mathrm{3}} +\:\left({n}+\mathrm{2}\:\right)^{\:\mathrm{3}} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 30/Jan/22
3(n+1) ∣ n^( 3) + (n+1)^( 3) +(n+2 )^( 3) _(−) ∵ n^( 3) + (n+1)^( 3) +(n+2 )^( 3) =(n+1−1)^3 + (n+1)^( 3) +(n+1+1 )^( 3) =(n+1)^3 −1−3(n+1)(n+1−1) +(n+1)^3 +(n+1)^3 +1+3(n+1)(n+1+1) =3(n+1)^3 −3n(n+1)+3(n+1)(n+2) =3(n+1){(n+1)^2 −3n+n+2} ∴ 3(n+1) ∣ n^( 3) + (n+1)^( 3) +(n+2 )^( 3)
$$\underset{−} {\:\mathrm{3}\left({n}+\mathrm{1}\right)\:\mid\:{n}^{\:\mathrm{3}} +\:\left({n}+\mathrm{1}\right)^{\:\mathrm{3}} +\left({n}+\mathrm{2}\:\right)^{\:\mathrm{3}} } \\ $$$$\because\:\:{n}^{\:\mathrm{3}} +\:\left({n}+\mathrm{1}\right)^{\:\mathrm{3}} +\left({n}+\mathrm{2}\:\right)^{\:\mathrm{3}} \\ $$$$\:=\left({n}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{3}} +\:\left({n}+\mathrm{1}\right)^{\:\mathrm{3}} +\left({n}+\mathrm{1}+\mathrm{1}\:\right)^{\:\mathrm{3}} \\ $$$$=\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}−\mathrm{3}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}−\mathrm{1}\right) \\ $$$$\:\:\:+\left({n}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:+\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}+\mathrm{3}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}+\mathrm{1}\right) \\ $$$$=\mathrm{3}\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{3}{n}\left({n}+\mathrm{1}\right)+\mathrm{3}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$=\mathrm{3}\left({n}+\mathrm{1}\right)\left\{\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}{n}+{n}+\mathrm{2}\right\} \\ $$$$\:\therefore\:\mathrm{3}\left({n}+\mathrm{1}\right)\:\mid\:{n}^{\:\mathrm{3}} +\:\left({n}+\mathrm{1}\right)^{\:\mathrm{3}} +\left({n}+\mathrm{2}\:\right)^{\:\mathrm{3}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 30/Jan/22
very nice solution sir ...
$$\:\:{very}\:{nice}\:{solution}\:{sir}\:… \\ $$
Answered by Rasheed.Sindhi last updated on 31/Jan/22
AnOther method 3(n+1) ∣ n^3 + (n+1)^3 + (n+2 )^3 _(−) ∵ n^3 + (n+1)^3 + (n+2 )^3 =a^3 +b^3 +c^3 −3abc+3abc ; { ((a=n)),((b=n+1)),((c=n+2)) :} =(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)+3abc =(n+n+1+n+2)( ∼ )+3n(n+1)(n+2) =3(n+1)( ∼ )+3n(n+1)(n+2) =3(n+1){( ∼ )+n(n+2)} ∴ 3(n+1) ∣ n^3 + (n+1)^3 +(n+2 )^3
$$\mathrm{AnOther}\:\mathrm{method} \\ $$$$\underset{−} {\:\mathrm{3}\left({n}+\mathrm{1}\right)\:\mid\:{n}^{\mathrm{3}} \:+\:\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\:\left({n}+\mathrm{2}\:\right)^{\mathrm{3}} } \\ $$$$\because\:\:{n}^{\mathrm{3}} \:+\:\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\:\left({n}+\mathrm{2}\:\right)^{\mathrm{3}} \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}+\mathrm{3}{abc}\:;\begin{cases}{{a}={n}}\\{{b}={n}+\mathrm{1}}\\{{c}={n}+\mathrm{2}}\end{cases} \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)+\mathrm{3}{abc} \\ $$$$=\left({n}+{n}+\mathrm{1}+{n}+\mathrm{2}\right)\left(\:\:\:\:\:\sim\:\:\:\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$=\mathrm{3}\left({n}+\mathrm{1}\right)\left(\:\sim\:\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$=\mathrm{3}\left({n}+\mathrm{1}\right)\left\{\left(\:\sim\:\right)+{n}\left({n}+\mathrm{2}\right)\right\} \\ $$$$ \\ $$$$\therefore\:\:\mathrm{3}\left({n}+\mathrm{1}\right)\:\mid\:{n}^{\mathrm{3}} +\:\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\left({n}+\mathrm{2}\:\right)^{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 31/Jan/22
thank you so much sir Rasheed
$$\:\:{thank}\:{you}\:{so}\:{much}\:{sir}\:\mathrm{R}{asheed} \\ $$

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