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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2019 · 8 April · Morning Shift · Mathematics · Binomial Theorem

The sum of the series 2 . ^ 20 C _ 0 + 5 . ^ 20 C _ 1 + 8 . ^ 20 C...

The sum of the series 2.20C0+5.20C1+8.20C2+11.20C3+.......+62.20C20{2 .}^{20}{C}_{0} + {5 .}^{20}{C}_{1} + {8 .}^{20}{C}_{2} + {11 .}^{20}{C}_{3} + . . . . . . . + {62 .}^{20}{C}_{20} is equal to

Options

  1. A226{2}^{26}
  2. B225{2}^{25}
  3. C224{2}^{24}
  4. D223{2}^{23}

Correct answer

B. 225{2}^{25}

Step-by-step solution

Let,

S=2.20C0+5.20C1+8.20C3+.......+62.20C20...S = 2{.}^{20}{C}_{0} + 5{.}^{20}{C}_{1} + 8{.}^{20}{C}_{3} + . . . . . . . + 62{.}^{20}{C}_{20} . . .\left(i\right)
S=62.20C0+59.20C1+56.20C3+.......+2.20C20...S = 62{.}^{20}{C}_{0} + 59{.}^{20}{C}_{1} + 56{.}^{20}{C}_{3} + . . . . . . . + 2{.}^{20}{C}_{20} . . .\left(ii\right)(Writinginreverseorder(Writing in reverse ordern{C}_{r}{=}^{n}{C}_{n - r})<br/>Adding(i)and(ii))<br />Adding \left(i\right) and \left(ii\right)
2S=64(20C0+20C1+20C2+.....+20C20)2S = 64{(}^{20}{C}_{0}{+}^{20}{C}_{1}{+}^{20}{C}_{2} + . . . . . {+}^{20}{C}_{20} )
⇒S=32.220=225⇒ S = {32.2}^{20} = {2}^{25}

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