# Question related with Binomial Theorem

#### happy21

How can we find the index 'n' of the binomial

$$\displaystyle \left ( \frac{x}{5}+\frac{2}{5} \right )^{n}$$, $$\displaystyle n\epsilon N$$

if the 9th term of the expansion has numerically the greatest coefficient.

Thx.

#### mathman

Forum Staff
The coefficients are $\binom{n}{k}$ for $0\le k\le n$. The maximum is in the middle. If $n$ is even then the max is at $\frac{n}{2}$. If $n$ is odd, it is max at a pair at $\frac{n-1}{2}$ and $\frac{n+1}{2}$.

You can work it out yourself. Be careful, $k$ starts at $0$.

• 1 person

#### happy21

The coefficients are $\binom{n}{k}$ for $0\le k\le n$. The maximum is in the middle. If $n$ is even then the max is at $\frac{n}{2}$. If $n$ is odd, it is max at a pair at $\frac{n-1}{2}$ and $\frac{n+1}{2}$.

You can work it out yourself. Be careful, $k$ starts at $0$.
But as given 9th term is the greatest (numerically), there should be (?) only one middle term. So 'n' should (or must?) be even? Further, what value(s) 'x' can take because it is also unknown in the question. Thx.

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#### SDK

The coefficients are $\binom{n}{k}$ for $0\le k\le n$. The maximum is in the middle. If $n$ is even then the max is at $\frac{n}{2}$. If $n$ is odd, it is max at a pair at $\frac{n-1}{2}$ and $\frac{n+1}{2}$.

You can work it out yourself. Be careful, $k$ starts at $0$.
This is not correct. The coefficient of $x^k$ is actually $2^{n-k} 5^{-n} \binom{n}{k}$.

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#### happy21

This is not correct. The coefficient of $x^k$ is actually $2^{n-k} 5^{-n} \binom{n}{k}$.
And after that?

#### mathman

Forum Staff
And after that?
Assuming he wants the coefficient of a polynomial in x, you are correct. I was assuming something different.