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Question-19060




Question Number 19060 by mondodotto@gmail.com last updated on 03/Aug/17
Commented by prakash jain last updated on 04/Aug/17
can u please use an app called  camscanner to take clear images  of handwritten or printed material.
$${can}\:{u}\:{please}\:{use}\:{an}\:{app}\:{called} \\ $$$${camscanner}\:{to}\:{take}\:{clear}\:{images} \\ $$$${of}\:{handwritten}\:{or}\:{printed}\:{material}. \\ $$
Commented by mondodotto@gmail.com last updated on 04/Aug/17
thAnKs
$$\mathrm{thAnKs} \\ $$
Answered by ajfour last updated on 04/Aug/17
(1+ax)^n =1+nax+((n(n−1)a^2 x^2 )/2)+...  given: (1+ax)^n =1+24x+24ax^2 +...  upon comparison:   na=24     and ((n(n−1)a^2 )/2)=24a  ⇒  na=24   and  ((na(n−1))/2)=24  As   na=24   ,   ((24(n−1))/2)=24  ⇒   ((n−1)/2)=1  ⇒    n−1=2     n=3  .  since  na=24   ⇒  a=((24)/n)  ⇒    a=((24)/3) =8 .
$$\left(\mathrm{1}+\mathrm{ax}\right)^{\mathrm{n}} =\mathrm{1}+\mathrm{nax}+\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+… \\ $$$$\mathrm{given}:\:\left(\mathrm{1}+\mathrm{ax}\right)^{\mathrm{n}} =\mathrm{1}+\mathrm{24x}+\mathrm{24ax}^{\mathrm{2}} +… \\ $$$$\mathrm{upon}\:\mathrm{comparison}: \\ $$$$\:\mathrm{na}=\mathrm{24}\:\:\:\:\:\mathrm{and}\:\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{24a} \\ $$$$\Rightarrow\:\:\mathrm{na}=\mathrm{24}\:\:\:\mathrm{and}\:\:\frac{\mathrm{na}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{24} \\ $$$$\mathrm{As}\:\:\:\mathrm{na}=\mathrm{24}\:\:\:,\:\:\:\frac{\mathrm{24}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{24} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\mathrm{n}−\mathrm{1}=\mathrm{2}\:\:\: \\ $$$$\mathrm{n}=\mathrm{3}\:\:. \\ $$$$\mathrm{since}\:\:\mathrm{na}=\mathrm{24}\:\:\:\Rightarrow\:\:\mathrm{a}=\frac{\mathrm{24}}{\mathrm{n}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{a}=\frac{\mathrm{24}}{\mathrm{3}}\:=\mathrm{8}\:. \\ $$
Commented by mondodotto@gmail.com last updated on 03/Aug/17
can you explain more please
$$\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{more}\:\mathrm{please} \\ $$$$ \\ $$

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