MAths Please PLEase PLease?

Question

Math Problems Please?
Hey, everyone these questions are due on wednesday, and ive done most of the sheet and i just have these questions left. Could someone please answer with full solutions. thank you so much in advance, and help will be appreciated, as the questions have been hard for me to solve and i really would like to know, how it should be done. Thank You... Or.. you could send me a e-mail to [e-mail], and i could send u the worksheet, only a few questions. thanks...

Question 17) Find the coefficient of:

a) x in (3x^2 + 2/x)^5

b) x in (x-2^x-1)^5

c) x^3 in (x-2/x)^7

d) x^3 in (x-1/2x)^9

e) x^4 in (x-1/2x)^10

f) p^4q^2 in (p+q)^6

g) x^5 in (1+2x)^7

h) x^3 in (2x-2)^10

i) x^5 in (x - 2/ x^2)^11

j) x^5 in (x+1/x)^9


Question 16) Find:

b) the fourth term in (1-x/2)^9

c) the fifth term in (1+x^2)^10

d) the sixth term in (3-2x)^8

e) the seventh term in (1-x/2)^10

THANKS HEAPS IN ADVANCE..

Answer 1

These questions are designed to give you
practice with the Binomial Formula, which is:
(x + y)^n = x^n + nx^(n-1)y + n(n-1)/2!*x^(n-2)y^2 + ...
Check it out in your textbook or Google it.
It helps to know Pascal's Triangle too.
I'll just do the first one to get you started.

Coefficient of x in (3x^2 + 2/x)^5.

(3x^2 + 2/x)^5

= (3x^2)^5 + 5(3x^2)^4(2/x) + 10(3x^2)^3(2/x)^2
+ 10(3x^2)^2(2/x)^3 + 5(3x^2)(2/x)^4 + (2/x)^5

= 243x^10 + 810x^7 + 1080x^4 + 720x + 240/x^2 + 32/x^5

Therefore, the coefficient of x is 720.

Of course, you don't have to work out the coefficient
of every term, just calculate all the x's, and then just
calculate the coefficient you need.
Good luck with that lot. You'll need plenty of time!

Answer 2

17a)
coefficient x in (3x^2 + 2/x)^5

no term involving x power 1 from the expansion

Answer 3

Send your sheet to me at [e-mail] I will solve your problems with full answers by late tomorrow afternoon. If you still need it.

Answer 4

the only way to learn is do it yourself..read the book!!!!