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anonymous
 4 years ago
find the coefficient of x^6 in the expansion of(5+2x^2)^7
anonymous
 4 years ago
find the coefficient of x^6 in the expansion of(5+2x^2)^7

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my memory fails me on this  ill go and check

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It uses Mimi's Crazy theorem ;)

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0\[t_{k} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{nk} b^{k} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mimi, that should be \( t_{k+1} \)

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0\[t_{6}= \left(\begin{matrix}7 \\ 6\end{matrix}\right)5\times(2x^2)^6\]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0Nope..its only finding the coeffient

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0Not the independent term i think..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You think wrong then :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its 7C3 5^4 2^3 = 35 * 625 * 8 = 175,000

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's crazy right Mimi? lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where came from 7c3 5^4 2^3

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0I think that im not wrong, k+1 is used for something else..cant remember..

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0& this is not crazy, the other part is crazy ::

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because you have 2x^2 in the paraenthesis it will be the fourth term in the expansion so formula is 7C3 * 5^(73) *2^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what this teory name?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i used the formula for the (r+1) th term of the binomial expansio (a+x)^n (r+1)th term = nC r a^(nr) x^r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mimi, I am sure that the general term of \( (a+b)^n \) is \( t_{k+1} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{nk} b^{k} \)

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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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