Question related with Binomial Theorem

Jan 2012
140
2
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
May 2007
6,933
775
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 $0$.
 
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Jan 2012
140
2
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.
 
Last edited:

SDK

Sep 2016
804
545
USA
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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