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Question Number 19083 by mondodotto@gmail.com last updated on 04/Aug/17

Commented by ajfour last updated on 04/Aug/17

something wrong with the question !

$$\mathrm{something}\:\mathrm{wrong}\:\mathrm{with}\:\mathrm{the}\:\mathrm{question}\:! \\ $$

Commented by mondodotto@gmail.com last updated on 04/Aug/17

nothing wrong

$$\mathrm{nothing}\:\mathrm{wrong} \\ $$

Commented by prakash jain last updated on 04/Aug/17

(2+ax)^n =n^2 +224x+1176x^2 +.. expression is true for all x x=0 n^2 =2^n n=2,4 n2^(n−1) ax=224x 2∙2^1 a=224⇒a=56 4.2^3 a=224⇒a=7 n=2 (1/2)n(n−1).2^(n−2) a^2 =56^2 =3136 n=4 6×2^2 ×49=2352=1176 n=4,a=7

$$\left(\mathrm{2}+{ax}\right)^{{n}} ={n}^{\mathrm{2}} +\mathrm{224}{x}+\mathrm{1176}{x}^{\mathrm{2}} +.. \\ $$$${expression}\:{is}\:{true}\:{for}\:{all}\:{x} \\ $$$${x}=\mathrm{0} \\ $$$${n}^{\mathrm{2}} =\mathrm{2}^{{n}} \\ $$$${n}=\mathrm{2},\mathrm{4} \\ $$$${n}\mathrm{2}^{{n}−\mathrm{1}} {ax}=\mathrm{224}{x} \\ $$$$\mathrm{2}\centerdot\mathrm{2}^{\mathrm{1}} {a}=\mathrm{224}\Rightarrow{a}=\mathrm{56} \\ $$$$\mathrm{4}.\mathrm{2}^{\mathrm{3}} {a}=\mathrm{224}\Rightarrow{a}=\mathrm{7} \\ $$$${n}=\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}−\mathrm{1}\right).\mathrm{2}^{{n}−\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{56}^{\mathrm{2}} =\mathrm{3136} \\ $$$${n}=\mathrm{4} \\ $$$$\mathrm{6}×\mathrm{2}^{\mathrm{2}} ×\mathrm{49}=\mathrm{2352}=\mathrm{1176} \\ $$$${n}=\mathrm{4},{a}=\mathrm{7} \\ $$

Commented by prakash jain last updated on 04/Aug/17

question probably should be n^2 +n+1=a((√n)+1)

$${question}\:{probably}\:{should}\:{be} \\ $$$${n}^{\mathrm{2}} +{n}+\mathrm{1}={a}\left(\sqrt{{n}}+\mathrm{1}\right) \\ $$

Answered by ajfour last updated on 04/Aug/17

(2+ax)^n =2^n +n(2^(n−1) )(ax) +((n(n−1))/2)(2^(n−2) )(a^2 x^2 )+.... and it is given that (2+ax)^n =n^2 +224x+1176x^2 +... comparing ((coefficient of x^0 )/(coefficient of x)) (2^n /(na2^(n−1) ))=(n^2 /(224)) ⇒ 2×224=n^3 a or n^3 a=448 ...(i) comparing ((coefficient of x)/(coefficient of x^2 )) ((na(2^(n−1) ))/({((na^2 (n−1)2^(n−2) )/2)}))=((224)/(1176)) ⇒ (4/((n−1)a)) = (4/(21)) ⇒ (n−1)a=21 ....(ii) dividing (i) by (ii) yields (n^3 /(n−1)) = ((448)/(21)) = ((64)/3) or 3n^3 =(n−1)(64) ⇒ n=4 using eqn. (ii) we find a=7 So, n^2 +n+a=16+4+7=27 while a((√n)+1) = 7(2+1)=21 . something is wrong...

$$\left(\mathrm{2}+\mathrm{ax}\right)^{\boldsymbol{\mathrm{n}}} =\mathrm{2}^{\boldsymbol{\mathrm{n}}} +\mathrm{n}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} \right)\left(\mathrm{ax}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}−\mathrm{2}} \right)\left(\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)+.... \\ $$$$\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{given}\:\mathrm{that} \\ $$$$\left(\mathrm{2}+\mathrm{ax}\right)^{\boldsymbol{\mathrm{n}}} =\mathrm{n}^{\mathrm{2}} +\mathrm{224x}+\mathrm{1176x}^{\mathrm{2}} +... \\ $$$$\mathrm{comparing}\:\:\:\frac{\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{0}} }{\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}} \\ $$$$\frac{\mathrm{2}^{\boldsymbol{\mathrm{n}}} }{\mathrm{na2}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} }=\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{224}}\:\:\:\Rightarrow\:\mathrm{2}×\mathrm{224}=\mathrm{n}^{\mathrm{3}} \mathrm{a}\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{or}\:\:\:\:\mathrm{n}^{\mathrm{3}} \mathrm{a}=\mathrm{448}\:\:\:\:\:...\left(\mathrm{i}\right) \\ $$$$\mathrm{comparing}\:\:\:\frac{\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}}{\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{na}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} \right)}{\left\{\frac{\mathrm{na}^{\mathrm{2}} \left(\mathrm{n}−\mathrm{1}\right)\mathrm{2}^{\boldsymbol{\mathrm{n}}−\mathrm{2}} }{\mathrm{2}}\right\}}=\frac{\mathrm{224}}{\mathrm{1176}} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{4}}{\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}}\:=\:\frac{\mathrm{4}}{\mathrm{21}}\:\: \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}=\mathrm{21}\:\:\:\:....\left(\mathrm{ii}\right) \\ $$$$\mathrm{dividing}\:\left(\mathrm{i}\right)\:\mathrm{by}\:\left(\mathrm{ii}\right)\:\mathrm{yields} \\ $$$$\:\:\:\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}−\mathrm{1}}\:=\:\frac{\mathrm{448}}{\mathrm{21}}\:=\:\frac{\mathrm{64}}{\mathrm{3}} \\ $$$$\mathrm{or}\:\:\:\:\:\:\:\:\mathrm{3n}^{\mathrm{3}} =\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{64}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:\:\:\mathrm{n}=\mathrm{4} \\ $$$$\:\:\mathrm{using}\:\mathrm{eqn}.\:\left(\mathrm{ii}\right)\:\mathrm{we}\:\mathrm{find} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}=\mathrm{7} \\ $$$$\mathrm{So},\:\:\:\:\:\boldsymbol{\mathrm{n}}^{\mathrm{2}} +\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{a}}=\mathrm{16}+\mathrm{4}+\mathrm{7}=\mathrm{27} \\ $$$$\mathrm{while}\:\:\:\:\boldsymbol{\mathrm{a}}\left(\sqrt{\boldsymbol{\mathrm{n}}}+\mathrm{1}\right)\:=\:\mathrm{7}\left(\mathrm{2}+\mathrm{1}\right)=\mathrm{21}\:. \\ $$$$\:\mathrm{something}\:\mathrm{is}\:\mathrm{wrong}... \\ $$$$ \\ $$

Commented by mondodotto@gmail.com last updated on 04/Aug/17

why 2^(n−1) and 2^(n−2) in your first step please recheck

$$\mathrm{why}\:\mathrm{2}^{\mathrm{n}−\mathrm{1}} \:\mathrm{and}\:\mathrm{2}^{\mathrm{n}−\mathrm{2}} \:\mathrm{in}\:\mathrm{your}\:\mathrm{first}\:\mathrm{step}\:\mathrm{please}\:\mathrm{recheck} \\ $$

Commented by ajfour last updated on 04/Aug/17

binomial theorem (x+y)^n =x^n +nx^(n−1) y+((n(n−1))/2)x^(n−2) y^2 +... .....+nxy^(n−1) +y^n .

$$\mathrm{binomial}\:\mathrm{theorem} \\ $$$$\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{n}} =\mathrm{x}^{\mathrm{n}} +\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \mathrm{y}+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}\mathrm{x}^{\mathrm{n}−\mathrm{2}} \mathrm{y}^{\mathrm{2}} +... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....+\mathrm{nxy}^{\mathrm{n}−\mathrm{1}} +\mathrm{y}^{\mathrm{n}} \:. \\ $$

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