Question

What is the 5th term of ( 2x-3y) ^10

Answer 1

\[\Large \text{Binomial Series:}\quad (a + b)^{n} = \sum_{k = 0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)a ^{n - k}b ^{k}\]

Answer 2

h:( i dont understand

Answer 3

\[\Large \left(\begin{matrix}n \\ k\end{matrix}\right) = \frac{ n! }{ (n-k)! k! }\]

Answer 4

\[\Large k^{th}\text{ term of Binomial Series:} \quad (a + b)^{n}= \left(\begin{matrix}n \\ k\end{matrix}\right)a ^{n - k}b ^{k}\]

Answer 5

i got the answer wrong because of that .. in my exam .. can you do it for me please that can i review

Answer 6

Just remember the kth term of the binomial series shown above and compare it to the problem.
a = 2x ; b = (-3y) ; n = 10 ; They are interested in the fifth term. K starts with 0 and therefore, k = 0, 1, 2, 3, 4, .... k = 4 is the 5th term.

Answer 7

how please :(

Answer 8

Just plug into the formula for the kth term:
\[\Large \left(\begin{matrix}10 \\ 4\end{matrix}\right)(2x)^{10-4}(-3y)^{4}\]

Answer 9

(10/4)?

Answer 10

No. \[\Large \left(\begin{matrix}n \\ k\end{matrix}\right) = \frac{ n! }{ (n-k)! k! }\]
\[\Large \left(\begin{matrix}10 \\ 4\end{matrix}\right) = \frac{ 10! }{ (10-4)! 4! }\]

Answer 11

thats the 5th term?

Answer 12

\[\Large = \frac{ 10*9*8*7 }{ 4*3*2*1 }\]

Answer 13

No, the whole thing shown earlier is the 5th term:\[\Large \left(\begin{matrix}10 \\ 4\end{matrix}\right)(2x)^{10-4}(-3y)^{4}\]

Answer 14

64x^6-81y^4?this is the 5th term?

Answer 15

You need to compute\[\Large \left(\begin{matrix}10 \\ 4\end{matrix}\right) = \frac{ 10*9*8*7 }{ 4*3*2*1 } = 210\]
So the 5th term is: 210 * (64x^6) * (-81y^4) = ? (simplify)

Answer 16

ok. wait

Answer 17

The 81 should not have a negative sign. since (-3)^4 = +81

Answer 18

* * ?

Answer 19

* means times or multiplication.

Answer 20

can draw it

Answer 21

^^^ what do you mean?

Answer 22

i tried but icnt

Answer 23

which part you are having trouble with?

Answer 24

It is a straightforward substitution once you remember the formula for the kth term of a binomial expansion.

Answer 25

13440x^6 + 81y^4? the answer

Answer 26

There is no addition in the middle. Each term has to be multiplied.
5th term is: 210 times (64x^6) times (81y^4) = ?

Answer 27

1088640x^6y^4?

Answer 28

yes.

Answer 29

how about this question ? what is the 7th term of (x-1/x)^11

Answer 30

a = x ; b = -1/x ; n = 11 ; k = 6 (7th term means k = 6 because k starts from 0)

Plug it into the formula.

Answer 31

help me please

Answer 32

\[\left(\begin{matrix}11 \\ 6\end{matrix}\right) = \frac{ 11! }{ 6!5! } = \frac{ 11*10*9*8*7 }{ 5*4*3*2*1 } = 462\]

Answer 33

a^(n-k) = (x)^(11-6) = x^5
b^k = (-1/x)^6 = +1/x^6

Multiply them altogether: 462 * x^5 * 1/x^6 = 462/x

Answer 34

462/x
?

Answer 35

yes.

Answer 36

where's the exponent?

Answer 37

1/x = x^(-1)

Because we have x and reciprocal x in: (x-1/x)^11
The powers of x and 1/x will partly cancel out in each term of the binomial expansion.

Answer 38

And lastly find the term that contains x^11 in the expansion of (3x-2/x^3)^15

Answer 39

Does x^3 divide just the 2 or the whole (3x-2) ?