Question

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Answer 1

Answer:

Step-by-step explanation:

According to the binomial theorem, we have,

$\left(a+b\right)^{n}={\,}^{n}C_{0}\,{a}^{n}+{\,}^{n}C_{1}\,{a}^{n-1}\,{b}+{\,}^{n}C_{2}\,{a}^{n-2}\,{b}^{2}+\cdots+{\,}^{n}C_{n}\,{b}^{n}$

To obtain the expression given in the question,

Put a = b = 1,

$\implies\left(1+1\right)^{n}={\,}^{n}C_{0}\,{(1)}^{n}+{\,}^{n}C_{1}\,{(1)}^{n-1}\,{(1)}+{\,}^{n}C_{2}\,{(1)}^{n-2}\,{(1)}^{2}+\cdots+{\,}^{n}C_{n}\,{(1)}^{n}$

$\implies\left(2\right)^{n}=1+{\,}^{n}C_{1}+{\,}^{n}C_{2}+\cdots+{\,}^{n}C_{n}$

$\implies\left(2\right)^{n}-1={\,}^{n}C_{1}+{\,}^{n}C_{2}+\cdots+{\,}^{n}C_{n}$

Answer 2

Question :-

$\rm \: What \: is \: C(n, 1) + \: C(n, 2) + \cdots + \: C(n, n) =$

$\rm \: (a) \: \: \: 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n}$

$\rm \: (b) \: \: \: 1 + 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n}$

$\rm \: (c) \: \: \: 1 + 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n - 1}$

$\rm \: (d) \: \: \: 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n - 1}$

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

We know that

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

can be further rewritten as

$\rm \: {(1 + x)}^{n} = \: 1 + \: C(n, 1)x + \: C(n, 2) {x}^{2} + \cdots \cdots + \: C(n, n) {x}^{n}$

On substituting x = 1, we get

$\rm \: {(1 + 1)}^{n} = \: 1 + \: C(n, 1) + \: C(n, 2) + \cdots \cdots + \: C(n, n)$

$\rm \: {2}^{n} = \: 1 + \: C(n, 1) + \: C(n, 2) + \cdots \cdots + \: C(n, n)$

$\rm\implies \: \: C(n, 1) + \: C(n, 2) + \cdots + \: C(n, n) = {2}^{n} - 1 - - (1)$

Now, Consider

$\rm \: 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n}$

Since, its a GP series with First term 2 and common ratio with n terms.

So, sum of n terms of GP series is

$\rm \: S_n \: = \: \frac{2( {2}^{n} - 1)}{2 - 1}$

$\rm\implies \:S_n = 2( {2}^{n} - 1) \ne C(n, 0) + C(n, 1) + ... + C(n, n)$

Now, Consider

$\rm \:1 + 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n}$

Since, its a GP series with First term 1 and common ratio with n + 1 terms.

So, sum of n + 1 terms of GP series is

$\rm \: S_{n + 1} \: = \: \frac{1( {2}^{n + 1} - 1)}{2 - 1}$

$\rm\implies \:S_{n + 1} = {2}^{n + 1} - 1 \ne C(n, 0) + C(n, 1) + ... + C(n, n)$

Now, Consider

$\rm \:1 + 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n - 1}$

Since, its a GP series with First term 1 and common ratio with n terms.

So, sum of n terms of GP series is

$\rm \: S_n \: = \: \frac{1( {2}^{n} - 1)}{2 - 1}$

$\rm\implies \:S_n = {2}^{n} - 1=C(n, 0) + C(n, 1) + ... + C(n, n)$

Now, Consider

$\rm \: 2 + {2}^{2} + {2}^{3} + \cdots \: + {2}^{n - 1}$

Since, its a GP series with First term 2 and common ratio with n - 1 terms.

So, sum of n - 1 terms of GP series is

$\rm \: S_{n - 1} \: = \: \frac{2( {2}^{n - 1} - 1)}{2 - 1}$

$\rm\implies \:S_{n - 1} = {2}^{n} - 2 \ne C(n, 0) + C(n, 1) + ... + C(n, n)$

So, from above calculations, we concluded that option (c) is correct.