Question Number 56075 by Kunal12588 last updated on 10/Mar/19
![Prove that If a set consist of n number of terms then its Power Set would contain 2^n number of terms. [Use formulas of sequence and series]](Q56075.png)
$${Prove}\:{that}\:{If}\:{a}\:{set}\:{consist}\:{of}\:{n}\:{number}\:{of} \\ $$$${terms}\:{then}\:{its}\:{Power}\:{Set}\:{would}\:{contain} \\ $$$$\mathrm{2}^{{n}} \:{number}\:{of}\:{terms}. \\ $$$$\left[{Use}\:{formulas}\:{of}\:{sequence}\:{and}\:{series}\right] \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Mar/19
![power set {∅}→nc_0 {a,b,c...}→nc_1 {ab,bc,cd...}→nc_2 ... .... nc_0 +nc_1 +nc_2 +...+nc_n =2^n [(1+x)^n =1+nc_1 x+nc_2 x^2 +...+nc_n x^n put x=1→2^n =nc_0 +nc_1 +...+nc_n ]](Q56080.png)
$${power}\:{set} \\ $$$$\left\{\emptyset\right\}\rightarrow{nc}_{\mathrm{0}} \\ $$$$ \\ $$$$\left\{{a},{b},{c}...\right\}\rightarrow{nc}_{\mathrm{1}} \\ $$$$\left\{{ab},{bc},{cd}...\right\}\rightarrow{nc}_{\mathrm{2}} \\ $$$$... \\ $$$$.... \\ $$$${nc}_{\mathrm{0}} +{nc}_{\mathrm{1}} +{nc}_{\mathrm{2}} +...+{nc}_{{n}} =\mathrm{2}^{{n}} \\ $$$$\left[\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nc}_{\mathrm{1}} {x}+{nc}_{\mathrm{2}} {x}^{\mathrm{2}} +...+{nc}_{{n}} {x}^{{n}} \right. \\ $$$$\left.{put}\:{x}=\mathrm{1}\rightarrow\mathrm{2}^{{n}} ={nc}_{\mathrm{0}} +{nc}_{\mathrm{1}} +...+{nc}_{{n}} \right] \\ $$