If {left( {2 + frac{x}{3}} right)^{55}}& | vedclass

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Question

Let $(r+1)^{	ext {th }}$ and $(r+2)^{	ext {th }}$ term has equal coefficient

$\left(2+\frac{x}{3}ight)^{55}=2^{55}\left(1+\frac{x}{6}ight)^{5

Vedclass · Accepted Answer

$8^{th}$ and $9^{th}$

Vedclass · Answer

Let $(r+1)^{	ext {th }}$ and $(r+2)^{	ext {th }}$ term has equal coefficient

$\left(2+\frac{x}{3}ight)^{55}=2^{55}\left(1+\frac{x}{6}ight)^{55}$

$(r+1)^{	ext {th }}$ term $=2^{55\, 55} \mathrm{C}_{r}\left(\frac{x}{6}ight)^{r}$

Coefficient of $x^{r}$ is $2^{55\, 55} \mathrm{C}_{r} \frac{1}{6^{r}}$

$(r+2)^{t h}$ term $=2^{55\,55} C_{r+1}\left(\frac{x}{6}ight)^{r+1}$

Coefficient of $x^{r+1}$ is $2^{55\, 55} \mathrm{C}_{r+1} \cdot \frac{1}{6^{r+1}}$

Both coefficients are equal

$2^{55\,55} C_{r} \frac{1}{6^{r}}=2^{55\, 55} \mathrm{C}_{r+1} \frac{1}{6^{r+1}}$

$\frac{1}{{r!55 - r!}} = \frac{1}{{r - 1!54 - r!}}.\frac{1}{6}$

$6(r+1)=55-r$

$6 r+6=55-r$

$7 r=49$

$r=7$

$(r+1)=8$

Coefficient of $8^{th}$ and $9^{th}$ terms are equal.

If ${\left( {2 + \frac{x}{3}} \right)^{55}}$ is expanded in the ascending powers of $x$ and the coefficients of powers of $x$ in two consecutive terms of the expansion are equal, then these terms are

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