If (2+ x3)^55 is expanded in the ascending powers of x and the coefficients of powers of x…

Question

If $\left(2+\dfrac x3\right)^{55}$ 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:

Accepted Answer

$\left(2+\dfrac x3\right)^{55}=2^{55}\left(1+\dfrac x6\right)^{55}$. The $r$th term has coefficient $2^{55}\binom{55}{r}\dfrac1{6^r}$ (using $r$ as term index starting at 0). Setting the $r$th and $(r+1)$th term coefficients equal:

$$\binom{55}{r}\frac1{6^r}=\binom{55}{r+1}\frac1{6^{r+1}}\quad\Rightarrow\quad 6(r+1)=55-r\quad\Rightarrow\quad7r=49\quad\Rightarrow\quad r=7$$

So the equal coefficients occur at the 7th and 8th terms.

Solutions · 1