Question

I need to find the middle term of this...

Answer 1

\[(2m+n ^{3})^{12}\]

Answer 2

Ive thought of using factorials, n will be 12 and r will be 5

Answer 3

so...i dont know, would that be correct?

Answer 4

0 1 2 3 4 5 6 7 8 9 10 11 12

6 is going to be r

Answer 5

yes osanseviero u can ans the question by solving it through fragments..:)

Answer 6

fragments i.e factorials.

Answer 7

but should r be the element I want -1: (6-1=5)

Answer 8

or it should be 6 ?

Answer 9

r = 6 so isnt it 12 C 6, so 12! / (6!6!) whatever that is

Answer 10

ok, thanks...

Answer 11

are you sure that r is not the number you want minus 1?

Answer 12

yeah r is not anything minus 1. r is the _th term in the sequence that you're trying to find the coefficient of. so the coefficient is nCr

Answer 13

the formula is n! / r!(n-r)!

Answer 14

So that would be \[924a ^{6}b ^{6}\]...obviously a will be 2m and b the other one..is that correct?

Answer 15

There's always Pacal's Triange . . . ;-)

Answer 16

ok...thanks :)