Question

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

Answer 1

my memory fails me on this - ill go and check

Answer 2

It uses Mimi's Crazy theorem ;)

Answer 3

\[t_{k} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k} \]

Answer 4

Mimi, that should be \( t_{k+1} \)

Answer 5

\[t_{6}= \left(\begin{matrix}7 \\ 6\end{matrix}\right)5\times(2x^2)^6\]

Answer 6

Nope..its only finding the coeffient

Answer 7

Not the independent term i think..

Answer 8

You think wrong then :P

Answer 9

its 7C3 5^4 2^3 = 35 * 625 * 8 = 175,000

Answer 10

It's crazy right Mimi? lol

Answer 11

where came from 7c3 5^4 2^3

Answer 12

I think that im not wrong, k+1 is used for something else..cant remember..

Answer 13

& this is not crazy, the other part is crazy ::

Answer 14

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

Answer 15

what this teory name?

Answer 16

Binomial theorem.

Answer 17

wait..im wrong.

Answer 18

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

Answer 19

Mimi, I am sure that the general term of \( (a+b)^n \) is \( t_{k+1} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k} \)

Answer 20

yup - thats it Fool

Answer 21

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