Question

If${}^{n} C_1, \: {}^{n} C_2$and[/tex] {}^{n} C_3[/tex]are in AP.
Then, find the value of$\left[ {}^{n} C_1 + {}^{n} C_2 + {}^{n} C_3 \right]$

Hints :—

1. r! = 1 \times 2 \times 3 \times .............. \times rr!=1×2×3×..............×r
2. \Large{ {}^{n} C_r = \frac{n!}{(n - r)! \times r!} }nCr​=(n−r)!×r!n!​

Correctly explained answer to be BRAINLIEST. ​

Answer 1

ⁿC₁  , ⁿC₂  & ⁿC₃  are in AP

=> 2 * ⁿC₂ = ⁿC₁  +  ⁿC₃

=> 2 * n!/((n-2)!2!)   =  n!/(n-1)!1!   + n!/(n-3)!3!

=> 2 * n(n-1) / 2  =  n   + n(n-1)(n-2)/6

=>  6n(n-1) = 6n + n(n-1)(n-2)

=> 6(n-1) = 6 + (n-1)(n-2)

=> 6n - 6 = 6 + n² - 3n + 2

=> n² - 9n + 14 = 0

=> n² - 2n - 7n + 14 = 0

=> n(n-2) - 7(n-2) = 0

=> (n-7)(n-2) = 0

=> n = 7     n can not be less than 3 so n = 2 is not possible

⁷C₁ + ⁷C₂ + ⁷C₃

= 7 + 21 + 35

= 63

hope this helps you.

pls mark as brainliest.

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

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

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

$= 7 + 21 + 35$

$= 63$