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June 20th, 2019, 11:28 AM   #1
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Question related with Binomial Theorem

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.
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June 20th, 2019, 12:59 PM   #2
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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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June 20th, 2019, 01:07 PM   #3
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Quote:
Originally Posted by mathman View Post
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 by happy21; June 20th, 2019 at 01:09 PM.
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June 20th, 2019, 03:39 PM   #4
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Quote:
Originally Posted by mathman View Post
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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June 20th, 2019, 11:34 PM   #5
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Quote:
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This is not correct. The coefficient of $x^k$ is actually $2^{n-k} 5^{-n} \binom{n}{k}$.
And after that?
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June 21st, 2019, 12:29 PM   #6
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Quote:
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And after that?
Assuming he wants the coefficient of a polynomial in x, you are correct. I was assuming something different.
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