Question

please answer this question​

Answer 1

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

We know that,

$\rm :\longmapsto\: {(1 + x)}^{n} = \: ^nC_ 0 + \: ^nC_ 1x + \: ^nC_ 2 {x}^{2} + - - - - + ^nC_ n {x}^{n}$

can also be rewritten as

$\rm :\longmapsto\: {(1 + x)}^{n} = \: C_ 0 + \: C_ 1x + \: C_ 2 {x}^{2} + - - - - + C_ n {x}^{n}$

We know,

$\boxed{ \sf{ \: ^nC_ 0 \: = \: C_ 0 = 1}}$

So, using this, we get

$\rm :\longmapsto\: {(1 + x)}^{n} = \: 1 + \: C_ 1x + \: C_ 2 {x}^{2} + - - - - + C_ n {x}^{n}$

Now, put x = 1, we get

$\rm :\longmapsto\: {(1 + 1)}^{n} = \: 1 + \: C_ 1(1) + \: C_ 2 {(1)}^{2} + - - - + C_ n {(1)}^{n}$

$\rm :\longmapsto\: {2}^{n} = \: 1 + \: C_ 1 + \: C_ 2 + - - - + C_ n$

$\bf :\longmapsto\: C_ 1 + \: C_ 2 + - - - + C_ n = {2}^{n} - 1$

Hence, Option (b) is correct.

Additional Information :-

Let's solve few more examples of same type!!!

Question :- Prove that

$\bf :\longmapsto\: C_ 0 - \: C_ 1 + C_ 2 - C_3 - - - + {( - 1)}^{n} C_ n = 0$

Solution :-

We know that,

$\rm :\longmapsto\: {(1 + x)}^{n} = \: ^nC_ 0 + \: ^nC_ 1x + \: ^nC_ 2 {x}^{2} + - - - - + ^nC_ n {x}^{n}$

can also be rewritten as

$\rm :\longmapsto\: {(1 + x)}^{n} = \: C_ 0 + \: C_ 1x + \: C_ 2 {x}^{2} + - - - - + C_ n {x}^{n}$

Now, put x = - 1, we get

$\rm :\longmapsto\: {(1 - 1)}^{n} = \: C_ 0 + \: C_ 1( - 1) + \: C_ 2 {( - 1)}^{2} + - - + C_ n {( - 1)}^{n}$

$\rm :\longmapsto\: {0}^{n} = \: C_ 0 - \: C_ 1 + \: C_ 2 + - - + C_ n {( - 1)}^{n}$

$\bf :\longmapsto\: \: C_ 0 - \: C_ 1 + \: C_ 2 + - - + C_ n {( - 1)}^{n} = 0$

Hence, Proved