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Question Number 80243 by mr W last updated on 01/Feb/20

if x+(1/x)=a (a∈R) find x^n +(1/x^n )=? (n∈N)

$${if}\:{x}+\frac{\mathrm{1}}{{x}}={a}\:\left({a}\in\mathbb{R}\right) \\ $$$${find}\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=? \\ $$$$\left({n}\in\mathbb{N}\right) \\ $$

Commented by TawaTawa last updated on 01/Feb/20

Sir i thought you solved this before with sir mind is power Q 78931

$$\mathrm{Sir}\:\mathrm{i}\:\mathrm{thought}\:\mathrm{you}\:\mathrm{solved}\:\mathrm{this}\:\mathrm{before}\:\mathrm{with}\:\mathrm{sir}\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power} \\ $$$$\mathrm{Q}\:\mathrm{78931} \\ $$

Commented by TawaTawa last updated on 01/Feb/20

Or you have another way sir

$$\mathrm{Or}\:\mathrm{you}\:\mathrm{have}\:\mathrm{another}\:\mathrm{way}\:\mathrm{sir} \\ $$

Commented by TawaTawa last updated on 01/Feb/20

Or you have another way sir

$$\mathrm{Or}\:\mathrm{you}\:\mathrm{have}\:\mathrm{another}\:\mathrm{way}\:\mathrm{sir} \\ $$

Commented by behi83417@gmail.com last updated on 01/Feb/20

sir mrW! what do think about this? a^n +b^n =(a+b)^n −nab(a+b)^(n−2) [n≥2]

$$\mathrm{sir}\:\mathrm{mrW}!\:\mathrm{what}\:\mathrm{do}\:\mathrm{think}\:\mathrm{about}\:\mathrm{this}? \\ $$$$\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{n}}} +\boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{n}}} =\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\boldsymbol{\mathrm{n}}} −\boldsymbol{\mathrm{nab}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\boldsymbol{\mathrm{n}}−\mathrm{2}} \:\:\:\left[\boldsymbol{\mathrm{n}}\geqslant\mathrm{2}\right] \\ $$

Commented by behi83417@gmail.com last updated on 01/Feb/20

i mean that: is this true? or can we prove it?

$$\mathrm{i}\:\mathrm{mean}\:\mathrm{that}:\:\mathrm{is}\:\mathrm{this}\:\mathrm{true}?\:\mathrm{or}\:\mathrm{can}\:\mathrm{we}\:\mathrm{prove} \\ $$$$\mathrm{it}? \\ $$

Commented by mr W last updated on 01/Feb/20

no sir, a^n +b^n =(a+b)^n −nab(a+b)^(n−2) [n≥2] is not true.

$${no}\:{sir}, \\ $$$$\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{n}}} +\boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{n}}} =\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\boldsymbol{\mathrm{n}}} −\boldsymbol{\mathrm{nab}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\boldsymbol{\mathrm{n}}−\mathrm{2}} \:\:\:\left[\boldsymbol{\mathrm{n}}\geqslant\mathrm{2}\right] \\ $$$${is}\:{not}\:{true}. \\ $$

Commented by behi83417@gmail.com last updated on 01/Feb/20

a^2 +b^2 =(a+b)^2 −2ab(a+b)^0 a^3 +b^3 =(a+b)^3 −3ab(a+b)^1 .....

$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{2ab}\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{0}} \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{1}} \\ $$$$..... \\ $$

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