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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

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

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

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

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

$${(2+ax)}^n=n^2+224x+1176x^2+..$$$$expressionistrueforallx$$$$x=0$$$$n^2=2^n$$$$n=2,4$$$$n{2}^{n-1}ax=224x$$$$2\cdot 2^{1}a=224\Rightarrow a=56$$$$4.2^{3}a=224\Rightarrow a=7$$$$n=2$$$$\frac{1}{2}n(n-1).2^{n-2}{a}^2={56}^2=3136$$$$n=4$$$$6\times 2^2\times 49=2352=1176$$$$n=4,a=7$$

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

$$questionprobablyshouldbe$$$$n^2+n+1=a(\sqrt{n}+1)$$

Answered by ajfour last updated on 04/Aug/17

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

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

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

Commented by ajfour last updated on 04/Aug/17

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

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