If the sum $S =^{20} C_{2} + {2.}^{20} C_{3} + {3.}^{20} C_{4} + . . . . . + {19.}^{20} C_{20}$ can be expressed as $1 + p . 2^{q}$ , then find the minimum value of $(p + q)$ where $p$ and $q ϵ N$ .

Open in App

Solution

$S = \sum_{r = 1}^{20} {(r - 1)}^{20} C_{r}$
$= \sum_{r = 1}^{20}^{20} C_{r} - \sum_{r = 1}^{20}^{20} C_{r}$
$= 20 \sum_{r = 1}^{20}^{19} C_{r - 1} - (2^{20} - 1)$
$= 20 (2^{19}) - (2^{20} - 1)$
$= 1 + (2^{19}) (20 - 2)$
$= 1 + 18 (2^{19})$
$\Rightarrow p = 18$
$\Rightarrow q = 19$
$\Rightarrow p + q = 37$ .

Suggest Corrections

Similar questions

c

y = c (1 - x^{2})

1

p / q

p, q ϵ N

(p + q)

\frac{p}{q}, 1.4 ¯ ¯¯¯¯ ¯ 73

p ϵ z, q ϵ N

\sqrt{x}

2 Q + 3 P = \frac{3 P - Q}{2 R}

R = - 2

Q = 1

z^{5} = 32

(z^{5} - 32) = (z - 2) (z^{2} - p z + 4) (z^{2} - q z + 4)

p^{2} - 2 q

Join BYJU'S Learning Program