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-

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

p o w e r s e t

{\emptyset} \to {n c}_{0}

{a, b, c \dots} \to {n c}_{1}

{a b, b c, c d \dots} \to {n c}_{2}

\dots

\dots .

{n c}_{0} + {n c}_{1} + {n c}_{2} + \dots + {n c}_{n} = 2^{n}

[{(1 + x)}^{n} = 1 + {n c}_{1} x + {n c}_{2} x^{2} + \dots + {n c}_{n} x^{n}

p u t x = 1 \to 2^{n} = {n c}_{0} + {n c}_{1} + \dots + {n c}_{n}]