Question

If ${}^{n} C_1, {}^{n} C_2$ and ${}^{n} C_3$ are in AP.Then, find the value of ${}^{n} C_1 + {}^{n} C_2 +  {}^{n} C_3$

Hints :—

1. $r! = 1 \times 2 \times 3 \times ......... r$

2. $\large{ {}^{n} C_r = \frac{n!}{(n - r)! \times r!} }$

Correctly explained answer to be BRAINLIEST. ​

Answer 1

$\huge\boxed{Answer}$

nC1 , nC2 & nC3 are in A.P.

So ,

2 × nC2 = nC1 + nC3 --------(1)

Now ,

nC1 + nC2 + nC3

nC2 + (nC1 + nC3)

From eq (1) -

nC2 + 2 × nC2

3 × nC2

We know

[ nCr = n! / r! (n - r)! ]

3 × n! / (2! × (n - 2)! )

3 × [ n(n - 1) (n - 2)! / 2 × (n-2)! ]

3 × n(n - 1) / 2

3n(n - 1) / 2

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1. Three terms a, b, c are said to be in Arithmetic Progression ( AP) if

$2b = a + c$

2.

$r! = 1 \times 2 \times 3 \times ......... \times r$

3.

$\large{ {}^{n} C_r = \frac{n!}{(n - r)! \times r!} }$

GIVEN

${}^{n} C_1, {}^{n} C_2 \: and \: {}^{n} C_3  \: are \: in \: AP$

TO DETERMINE

${}^{n} C_1 + {}^{n} C_2 +  {}^{n} C_3$

EVALUATION

Since

${}^{n} C_1, {}^{n} C_2 \: and \: {}^{n} C_3  \: are \: in \: AP$

So

$2 \times {}^{n} C_2 = \: {}^{n} C_1 \: + \: {}^{n} C_3$

$\displaystyle \: 2 \times \frac{n!}{(n - 2)! \times 2!} = \frac{n!}{(n - 1)! \times 1!} + \frac{n!}{(n - 3)! \times 3!}$

$\displaystyle \: \implies \: 2 \times \frac{n(n - 1)}{2} = n + \frac{n(n - 1)(n - 2)}{6}$

$\displaystyle \: \implies \: (n - 1) = 1 + \frac{(n - 1)(n - 2)}{6} \: ( \because \: n \ne \: 0)$

$\implies \: 6n - 6 = 6 + {n}^{2} - 3n + 2$

$\implies \: {n}^{2} - 9n + 14 = 0$

$\implies \: {n}^{2} - 2n - 7n+ 14 = 0$

$\implies \: n(n - 2) - 7(n - 2) = 0$

$\implies \: (n - 2) (n - 7) = 0$

So

$either \: (n - 2) = 0 \: \: or \: \: (n - 7) = 0$

$so \: n \: = 2 \: \: or \: n \: = 7$

$since \: \: {}^{n} C_3 \: \: exists \: \: so \: \: n \: \geqslant 3$

So

$n \ne \: 2$

So

$n = 7$

RESULT

${}^{n} C_1 + {}^{n} C_2 +  {}^{n} C_3$

$= {}^{7} C_1 + {}^{7} C_2 +  {}^{7} C_3$

$= 7 + 21 + </u></em></strong><strong><em><u>35</u></em></strong><strong><em><u>$

$= </u></em></strong><strong><em><u>63</u></em></strong><strong><em><u>$