If ( 20 ) 19 + 2 ( 21 ) ( 20 ) 18 + 3 ( 21 ) 2 ( 20 ) 17 + . . . . + 20 ( 21 ) 19 = k ( 20 ) 19 , then k is equal to _________ . [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If ${(20)}^{19} + 2 (21) {(20)}^{18} + 3 {(21)}^{2} {(20)}^{17} + . . . . + 20 {(21)}^{19} = k {(20)}^{19},$ then $k$ is equal to _________ . [2023]

Ans.
(400)

We have,

${(20)}^{19} k = {(20)}^{19} + 2 \cdot 21 \cdot {(20)}^{18} + 3 \cdot {(21)}^{2} \cdot {(20)}^{17} + \dots + 20 \cdot {(21)}^{19}$

$\Rightarrow k {(20)}^{19} = 20^{19} [1 + 2 \cdot (\frac{21}{20}) + 3 \cdot {(\frac{21}{20})}^{2} + \dots + 20 \cdot {(\frac{21}{20})}^{19}]$

$\Rightarrow k = 1 + 2 \cdot \frac{21}{20} + 3 \cdot {(\frac{21}{20})}^{2} + \dots + 20 {(\frac{21}{20})}^{19} ...(i)$

$\Rightarrow \frac{21}{20} k = \frac{21}{20} + 2 \cdot {(\frac{21}{20})}^{2} + \dots + 19 \cdot {(\frac{21}{20})}^{19} + 20 \cdot {(\frac{21}{20})}^{20} ...(ii)$

$Subtracting (ii) from (i), we get$

$- \frac{k}{20} = 1 + (\frac{21}{20}) + {(\frac{21}{20})}^{2} + \dots + {(\frac{21}{20})}^{19} - 20 \cdot {(\frac{21}{20})}^{20}$

$= 1 {\frac{{(\frac{21}{20})}^{20} - 1}{\frac{21}{20} - 1}} - 20 \cdot {(\frac{21}{20})}^{20}$ $= 20 \cdot {(\frac{21}{20})}^{20} - 20 - 20 \cdot {(\frac{21}{20})}^{20}$

$\Rightarrow \frac{- k}{20} = - 20 \Rightarrow k = 400$

Want Us to Email you About Special Offers & Updates?