First slide

Arithmetico-Geometric Progression

Question

If ${(20)}^{19} + 2 (21) {(20)}^{18} + 3 {(21)}^{2} {(20)}^{17} + . . . + 20 {(21)}^{19} = k {(20)}^{19}$
then $k$ is equal to

Moderate

A

400
B

100
C

441
D

420

Solution

Dividing by ${(20)}^{19},$ we can write
   $k = 1 + 2 x + 3 x^{2} + \dots + 20 x^{19}$ where $x = 21 / 20 .$
$\begin{array}{l} k = \frac{d}{d x} [x + x^{2} + \dots + x^{20}] \\ = \frac{d}{d x} [\frac{x (x^{20} - 1)}{x - 1}] \\ = \frac{(x - 1) (21 x^{20} - 1) - x (x^{20} - 1)}{(x - 1)^{2}} \\ = \frac{20 x^{21} - 21 x^{20} + 1}{(x - 1)^{2}} \end{array}$
But $20 x^{21} - 21 x^{20}$
   $= 20 {(\frac{21}{20})}^{21} - 21 {(\frac{21}{20})}^{21} = 0$
$∴$    $k = \frac{1}{(21 / 20 - 1)^{2}} = 400$

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