In the following expansion find the value of the greatest term:
${(1 + x)}^{n}$ when $x = \frac{2}{3}, n = 6$ .

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Solution

$T_{r + 1} = \frac{7 - r}{r} \cdot \frac{2}{3} \times T_{r}$

For $T_{r + 1} > T_{r}$ when $14 - 2 r > 3 r$ or $r < \frac{14}{5}$

Therefore $T_{3}$ is the greatest terms and its value $= \frac{6 \cdot 5}{1 \cdot 2} \cdot {(\frac{2}{3})}^{2} = \frac{20}{9}$

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In the following expansion find the value of the greatest term:(1+x)n when x=23,n=6.

Tr+1=7−rr⋅23×TrFor Tr+1>Tr when 14−2r>3r or r<145Therefore T3 is the greatest terms and its value =6⋅51⋅2⋅(23)2=209

In the following expansion find the value of the greatest term:
${(1 + x)}^{n}$ when $x = \frac{2}{3}, n = 6$ .

$T_{r + 1} = \frac{7 - r}{r} \cdot \frac{2}{3} \times T_{r}$

For $T_{r + 1} > T_{r}$ when $14 - 2 r > 3 r$ or $r < \frac{14}{5}$

Therefore $T_{3}$ is the greatest terms and its value $= \frac{6 \cdot 5}{1 \cdot 2} \cdot {(\frac{2}{3})}^{2} = \frac{20}{9}$