Is the greatest term in the expansion of 1-1320

The correct option is C 8th nbsp;termTo nbsp;find nbsp;the nbsp;greatest nbsp;term nbsp;Tr+1Tr≥1Cr20(1/3)rCr 120(1/3)r 1≥1(20 r+1)r.13≥121 r≥3r3+1r≤ ...

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Byju's Answer

Standard XII

Mathematics

Geometric Progression

is the greate...

Question

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

6^{t h} term

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7^{t h} term

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8^{t h} term

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Solution

The correct option is C $8^{t h} term$
$To find the greatest term \frac{T_{r + 1}}{T_{r}} \geq 1 \frac{20 C r {(1 / \sqrt{3})}^{r}}{20 C r - 1 {(1 / \sqrt{3})}^{r - 1}} \geq 1 \frac{(20 - r + 1)}{r} . \frac{1}{\sqrt{3}} \geq 1 21 - r \geq \sqrt{3} r (\sqrt{3} + 1) r \leq 21 r \leq \frac{21}{(\sqrt{3} + 1)} r \leq 7.686 Thus, greatest term obtained for r = 7 So, T_{7 + 1} = T_{8} is numerically largest . 8^{th} term is the greatest .$

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

The correct option is C 8th termTo find the greatest term Tr+1Tr≥120Cr(1/√3)r20Cr−1(1/√3)r−1≥1(20−r+1)r.1√3≥121−r≥√3r(√3+1)r≤21r≤21(√3+1)r≤7.686Thus, greatest term obtained for r=7So, T7+1=T8 is numerically largest.8th term is the greatest.

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