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BITSAT 2018

Exam: BITSAT
Year: 2018
Subject: Mathematics
Chapter: Binomial Theorem

BITSAT 2018 · Mathematics · Binomial Theorem

The greatest term in the expansion of 3 (1 + (1)/(3))20 is

The greatest term in the expansion of

\sqrt{3}\left(1 + \dfrac{1}{3}\right)20

is

Options

A $\dfrac{26840}{9}$
B $\dfrac{24840}{9}$
C $\dfrac{25840}{9}$
DNone of these

Correct answer

C. $\dfrac{25840}{9}$

Step-by-step solution

${T}_{r + 1} = \sqrt{3}{·}^{20}{C}_{r} \left(\dfrac{1}{3}\right)r$

and~ ${T}_{r} = \sqrt{3}{·}^{20}{C}_{r - 1} \left(\dfrac{1}{3}\right)r - 1$

Now,~ $\dfrac{Tr + 1}{Tr} = \dfrac{20 - r + 1}{r}\left(\dfrac{1}{3}\right)$

Since,~ ${T}_{r + 1}$ \ge {T}_{r}

$⇒ 20 - r + 1 ≥ \sqrt{3}r$

$⇒ r ≤ \dfrac{21}{\sqrt{3} + 1} = \dfrac{21}{2 . 73}$

$⇒ r ≤ 7 . 692 ⇒ r = 7$

$∴$ The greatest term is ${T}_{8} = \sqrt{3} \cdotC7 20 \left(\dfrac{1}{3}\right)$ 7 = $\dfrac{25840}{9}</p>$

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