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

T r + 1 = 3 · 20 C r 1 3 r and T r = 3 · 20 C r - 1 1 3 r - 1 Now, T r + 1 T r = 20 - r + 1 r 1 3 Since, T r + 1 ≥ T r ⇒ 20 - r + 1 ≥ 3 r ⇒ r ≤ 21 3 + 1 = 21 2 . 73 ⇒ r ≤ 7 . 692 ⇒ r = 7 ∴ The greatest term is T 8 = 3 · C 7 20 1 3 7 = 25840 9

Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremGeneralMedium2 minPYQ_2018

MathematicsMediumsingle choice

The greatest term in the expansion of $\sqrt{3} {(1 + \frac{1}{\sqrt{3}})}^{20}$ is

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Answer:

Solution:

$T_{r + 1} = \sqrt{3} \cdot^{20} C_{r} {(\frac{1}{\sqrt{3}})}^{r}$

and $T_{r} = \sqrt{3} \cdot^{20} C_{r - 1} {(\frac{1}{\sqrt{3}})}^{r - 1}$

Now, $\frac{T_{r + 1}}{T_{r}} = \frac{20 - r + 1}{r} (\frac{1}{\sqrt{3}})$

Since, $T_{r + 1} \geq T_{r}$

$\Rightarrow 20 - r + 1 \geq \sqrt{3} r$

$\Rightarrow r \leq \frac{21}{\sqrt{3} + 1} = \frac{21}{2.73}$

$\Rightarrow r \leq 7.692 \Rightarrow r = 7$

$∴$ The greatest term is $T_{8} = \sqrt{3} \cdot {}^{20}C_{7} {(\frac{1}{\sqrt{3}})}^{7} = \frac{25840}{9}$

Stream:BITSATSubject:MathematicsTopic:Binomial TheoremSubtopic:General

⏱ 2mℹ️ Source: PYQ_2018

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