If (20) ^ 19 + 2(21) (20) ^ 18 + 3 (21) ^ 2 (20) ^ 17 + . . . + 20 (21) ^ 19 = k (20) ^ 19 , then k is equal to _____.

Correct answer: 0 Let, S = (20) ^ 19 + 2(21) (20) ^ 18 + 3 (21) ^ 2 (20) ^ 17 + . . . + 20 (21) ^ 19 ⇒ S = 20 ^ 19 + 2 · (20) ^ 19 · ((21)/(20)) + 3 (20) ^ 19 ((21)/(20)) ^ 2 + . . . + 20 (20) ^ 19 · ((21)/(20)) ^ 19 ⇒ S = 20 ^ 19 (1 + 2 · (21)/(20) + 3 ((21)/(20)) ^ 2 + . . . + 20 ((21)/(20)) ^ 19 ) Now on comparing with S = k · (20) ^ 19 we get, ⇒ k = 1 + 2 · ((21)/(20)) + 3 ((21)/(20)) ^ 2 + . . . . + 20 ((21)/(20)) ^ 19 . . . . (1) ⇒ (21)/(20)k = (21)/(20) + 2 ((21)/(20)) ^ 2 + . . . + 19 ((21)/(20)) ^ 19 + 20 ((21)/(20)) ^ 20 . . . . (2) Now on subtracting above equations we get, ⇒ k - (21)/(20)k = 1 + (21)/(20) + ((21)/(20)) ^ 2 + . . . + ((21)/(20)) ^ 19 - 20 ((21)/(20)) ^ 20 ⇒ (- k)/(20) = ((21)/(20)) ^ 20 - 1 (21)/(20) - 1 - 20 ((21)/(20)) ^ 20 ⇒ - (k)/(20) = ( ((21)/(20)) ^ 20 - 1) × 20 - 20 ((21)/(20)) ^ 20 ⇒ (- k)/(20) = - 20 ⇒ k = 400

JEE Main 20236 AprilEvening Shift

Exam: JEE Main
Year: 2023
Date: 6 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2023 · 6 April · Evening Shift · Mathematics · Sequences and Series

If (20) ^ 19 + 2(21) (20) ^ 18 + 3 (21) ^ 2 (20) ^ 17 + . . . + 20...

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

Correct answer

Step-by-step solution

Let,
$S = {\left(20\right)}^{19} + 2\left(21\right){\left(20\right)}^{18} + 3{\left(21\right)}^{2}{\left(20\right)}^{17} + . . . + 20{\left(21\right)}^{19}$
$⇒ S = {20}^{19} + 2 \cdot {\left(20\right)}^{19} \cdot \left(\dfrac{21}{20}\right) + 3{\left(20\right)}^{19}{\left(\dfrac{21}{20}\right)}^{2} + . . . + 20{\left(20\right)}^{19} \cdot {\left(\dfrac{21}{20}\right)}^{19}$
$⇒ S = {20}^{19}\left(1 + 2 \cdot \dfrac{21}{20} + 3{\left(\dfrac{21}{20}\right)}^{2} + . . . + 20{\left(\dfrac{21}{20}\right)}^{19}\right)$
Now on comparing with $S = k \cdot {\left(20\right)}^{19}$ we get,
$⇒ k = 1 + 2 \cdot \left(\dfrac{21}{20}\right) + 3{\left(\dfrac{21}{20}\right)}^{2} + . . . . + 20{\left(\dfrac{21}{20}\right)}^{19} . . . . \left(1\right)$
$⇒ \dfrac{21}{20}k = \dfrac{21}{20} + 2{\left(\dfrac{21}{20}\right)}^{2} + . . . + 19{\left(\dfrac{21}{20}\right)}^{19} + 20{\left(\dfrac{21}{20}\right)}^{20} . . . . \left(2\right)$
Now on subtracting above equations we get,
$⇒ k - \dfrac{21}{20}k = 1 + \dfrac{21}{20} + {\left(\dfrac{21}{20}\right)}^{2} + . . . + {\left(\dfrac{21}{20}\right)}^{19} - 20{\left(\dfrac{21}{20}\right)}^{20}$
$⇒ \dfrac{- k}{20} = \dfrac{{\left(\dfrac{21}{20}\right)}^{20} - 1}{\dfrac{21}{20} - 1} - 20{\left(\dfrac{21}{20}\right)}^{20}$
$⇒ - \dfrac{k}{20} = \left({\left(\dfrac{21}{20}\right)}^{20} - 1\right) \times 20 - 20{\left(\dfrac{21}{20}\right)}^{20}$
$⇒ \dfrac{- k}{20} = - 20$
$⇒ k = 400$

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