Question

$\textsf{Mathematically prove that :-}$

${^nC_0} + {^nC_1}+{^nC_2}+{^nC_3} +...+{^nC_n} = 2^n$

Solve this without using induction.
​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

The binomial expansion of

$\rm :\longmapsto\: {(x + y)}^{n}, \: where \: n \: is \: a \: positive \: integer$

is given as

$\rm \:  =  \:^nC_0 {x}^{n - 0} {y}^{0} + \: ^nC_1 {x}^{n - 1} {y}^{1} + \: ^nC_2 {x}^{n - 2} {y}^{2} + - - - + \: ^nC_n {x}^{0} {y}^{n}$

$\rm \:  =  \:^nC_0 {x}^{n} + \: ^nC_1 {x}^{n - 1} {y}^{1} + \: ^nC_2 {x}^{n - 2} {y}^{2} + - - - + \: ^nC_n{y}^{n}$

So, Put y = 1, in the above expansion, we get

$\rm :\longmapsto\: {(x + 1)}^{n}$

$\rm \:  =  \:^nC_0 {x}^{n} + \: ^nC_1 {x}^{n - 1} + \: ^nC_2 {x}^{n - 2} + - - - + \: ^nC_n$

Now, Put x = 1, in this expansion, we get

$\rm \: {(1 + 1)}^{n} = {^nC_0} {(1)}^{n} + {^nC_1} {(1)}^{n - 1} +{^nC_2} {(1)}^{n - 2} +{^nC_3} {(1)}^{n - 3} +...+{^nC_n}$

$\rm :\longmapsto\: {2}^{n} \: = \: {^nC_0} + {^nC_1}+{^nC_2}+{^nC_3} +...+{^nC_n}$

Hence, Proved

Additional Information :-

$\boxed{ \rm \:{^nC_0} + {^nC_2}+{^nC_4} + - - = +{^nC_1}+{^nC_3} + - - - = {2}^{n - 1}}$

$\boxed{ \rm \:^nC_x + \: ^nC_{x + 1} = \: ^{n + 1}C_x}$

$\boxed{ \rm \: \frac{^nC_x}{^nC_{x - 1}} = \frac{n - x + 1}{x}}$

$\boxed{ \rm \:^nC_x = \frac{n}{x} \: ^{n - 1}C_{x - 1}}$