For some n10, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion…

Question

For some $n
e10$, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion of $(1+x)^{n+4}$ be in A.P. Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is:

Accepted Answer

Coefficients of the 5th, 6th, 7th terms are $\binom{n+4}{4},\binom{n+4}{5},\binom{n+4}{6}$. For an A.P.:

$$2\binom{n+4}{5} = \binom{n+4}{4}+\binom{n+4}{6}$$

Using Pascal's identity repeatedly: $4\binom{n+4}{5}=\binom{n+5}{5}+\binom{n+5}{6}$, which simplifies (via factorial expansion) to

$$4=\frac{(n+6)(n+5)}{6n}\quad\Rightarrow\quad n^2-13n+30=0\quad\Rightarrow\quad n=3\text{ or }n=10$$

Since $n\ne10$, $n=3$, so $n+4=7$. The largest (middle) coefficient in $(1+x)^7$ is $\binom{7}{3}=\binom{7}{4}=35$.

Solutions · 1