Let the coefficients of three consecutive terms in the binomial expansion of (1+2x)^n be i…

Question

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio $2:5:8$. Then the coefficient of the term which is in the middle of these three terms is \_\_\_\_\_\_\_.

Answer: 1120

Accepted Answer

With $t_{r+1}={}^nC_r(2x)^r$: $\dfrac{{}^nC_{r-1}2^{r-1}}{{}^nC_r2^r}=\dfrac25\Rightarrow\dfrac{r}{n-r+1}=\dfrac45\Rightarrow9r=4(n+1)\quad\text{...(1)}$

$$\frac{{}^nC_r2^r}{{}^nC_{r+1}2^{r+1}}=\frac58\quad\Rightarrow\quad\frac{r+1}{n-r}=\frac54\quad\Rightarrow\quad5n-4=9r\quad\text{...(2)}$$

From (1) and (2): $4n+4=5n-4\Rightarrow n=8$, and from (1): $r=4$.

Middle coefficient: $\binom{8}{4}2^4=70\times16=1120$

Solutions · 1