7.Binomial Theorem
hard

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

A

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

B

$7^{th}$ and $8^{th}$

C

$28^{th}$ and $29^{th}$

D

$27^{th}$ and $28^{th}$

(JEE MAIN-2014)

Solution

Let $(r+1)^{\text {th }}$ and $(r+2)^{\text {th }}$ term has equal coefficient

$\left(2+\frac{x}{3}\right)^{55}=2^{55}\left(1+\frac{x}{6}\right)^{55}$

$(r+1)^{\text {th }}$ term $=2^{55\, 55} \mathrm{C}_{r}\left(\frac{x}{6}\right)^{r}$

Coefficient of $x^{r}$ is $2^{55\, 55} \mathrm{C}_{r} \frac{1}{6^{r}}$

$(r+2)^{t h}$ term $=2^{55\,55} C_{r+1}\left(\frac{x}{6}\right)^{r+1}$

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.

Standard 11
Mathematics

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