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]


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Question Number 56075 by Kunal12588 last updated on 10/Mar/19

$P r o v e t h a t I f a s e t c o n s i s t o f n n u m b e r o f$ $t e r m s t h e n i t s P o w e r S e t w o u l d c o n t a i n$ $2^{n} n u m b e r o f t e r m s .$ $[U s e f o r m u l a s o f s e q u e n c e a n d s e r i e s]$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Mar/19
	$power set {∅}→nc_0 {a,b,c...}→nc_1 {ab,bc,cd...}→nc_2 ... .... nc_0 +nc_1 +nc_2 +...+nc_n =2^n [(1+x)^n =1+nc_1 x+nc_2 x^2 +...+nc_n x^n put x=1→2^n =nc_0 +nc_1 +...+nc_n ]$
	$p o w e r s e t$ ${\emptyset} \to {n c}_{0}$ ${a, b, c . . .} \to {n c}_{1}$ ${a b, b c, c d . . .} \to {n c}_{2}$ $. . .$ $. . . .$ ${n c}_{0} + {n c}_{1} + {n c}_{2} + . . . + {n c}_{n} = 2^{n}$ $[{(1 + x)}^{n} = 1 + {n c}_{1} x + {n c}_{2} x^{2} + . . . + {n c}_{n} x^{n}$ $p u t x = 1 \to 2^{n} = {n c}_{0} + {n c}_{1} + . . . + {n c}_{n}]$
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