find the coefficient of x^6 in the expansion of(5+2x^2)^7

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find the coefficient of x^6 in the expansion of(5+2x^2)^7

Mathematics
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my memory fails me on this - ill go and check
It uses Mimi's Crazy theorem ;)
\[t_{k} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k} \]

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Mimi, that should be \( t_{k+1} \)
\[t_{6}= \left(\begin{matrix}7 \\ 6\end{matrix}\right)5\times(2x^2)^6\]
Nope..its only finding the coeffient
Not the independent term i think..
You think wrong then :P
its 7C3 5^4 2^3 = 35 * 625 * 8 = 175,000
It's crazy right Mimi? lol
where came from 7c3 5^4 2^3
I think that im not wrong, k+1 is used for something else..cant remember..
& this is not crazy, the other part is crazy ::
because you have 2x^2 in the paraenthesis it will be the fourth term in the expansion so formula is 7C3 * 5^(7-3) *2^3
what this teory name?
Binomial theorem.
wait..im wrong.
i used the formula for the (r+1) th term of the binomial expansio (a+x)^n (r+1)th term = nC r a^(n-r) x^r
Mimi, I am sure that the general term of \( (a+b)^n \) is \( t_{k+1} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k} \)
yup - thats it Fool
Well, in my book it says.. "Specially for the expansion of (a+b)^n , the k+1th term is...that thing..cbb typing it again

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